https://wiki.ubc.ca/api.php?action=feedcontributions&feedformat=atom&user=NarissarasuUBC Wiki - User contributions [en]2026-10-10T07:55:53ZUser contributionsMediaWiki 1.43.10https://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_13&diff=74826Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 132011-02-03T06:44:30Z<p>Narissarasu: </p>
<hr />
<div>===Brightness of Stars on a logarithmic scale===<br />
<br />
When observing stars through the naked eye we notice that some are brighter than others. But is it possible to determine how bright a star is based upon a mathematical denotation. Lets read on... :)<br />
<br />
When determining star brightness, we must find a scale into which compare them to. Stars are placed on a 1-6 magnitude scale in terms of brightness (1 being the brightest, and 6 being the least brightest). The difference in magnitude from 1 to 2 is <math> \sqrt[5]{100}=2.512</math>. The reason why we use this number is because the difference in terms of brightness from 1-6 magnitude 1 is 100x greater than in 6<br />
<br />
In order to continue our explanation the logarithmic scale of the brightness of stars, we will calculate the variation of brightness between the planet Neptune and the planet Venus. There are three necessary factors: average brightness of each planet, ratio of brightness and the variation of brightness.<br />
<br />
=====Average Brightness=====<br />
Neptune: 7.9<br />
Venus: -4.355<br />
<br />
=====Ratio of brightness=====<br />
<p>x = mf - mb</p><br />
<p>Mf = magnitude of the fainter planet</p><br />
<p>Mb = magnitude of the brighter planet</p><br />
<p>http://i.imgur.com/amuUv.png</p><br />
<p> The scale of the brightness, 1 being the brightest and 6 being the least bright for the star. The ratio of the brightness scale can be seen above as I,II,III,IV, and V</p><br />
<br />
=====Variation of brightness=====<br />
<p><math>Vb = 2.512^x</math></p><br />
<br />
<p>To be able to find the variation in brightness, you first need to find the ''difference in magnitude'' by using the equation for the ratio of brightness.</p><br />
<p>x = mf - mb</p><br />
<p>x = 7.9 - -4.355</p><br />
<p>x = 12.255</p><br />
<br />
<p> Then, plug in the the ratio of brightness (x = 12.255) into the variation of brightness. </p><br />
<p><math>Vb = 2.512^x</math></p><br />
<p><math>Vb = 2.512^{12.255}</math></p><br />
<p>Vb = 79843.695</p><br />
<br />
<p>Therefore, Venus is 7843.695 times brighter than Neptune.</p><br />
<br />
===!!!HEY TEAM SEND ME THE PIC YOU WANT ME TO CROP ONTO THE STAR!!!===<br />
MY EMAIL IS magasparian@gmail.com</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_13&diff=74824Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 132011-02-03T06:42:05Z<p>Narissarasu: </p>
<hr />
<div>===Brightness of Stars on a logarithmic scale===<br />
<br />
When observing stars through the naked eye we notice that some are brighter than others. But is it possible to determine how bright a star is based upon a mathematical denotation. Lets read on... :)<br />
<br />
When determining star brightness, we must find a scale into which compare them to. Stars are placed on a 1-6 magnitude scale in terms of brightness (1 being the brightest, and 6 being the least brightest). The difference in magnitude from 1 to 2 is <math> \sqrt[5]{100}=2.512</math>. The reason why we use this number is because the difference in terms of brightness from 1-6 magnitude 1 is 100x greater than in 6<br />
<br />
In order to continue our explanation the logarithmic scale of the brightness of stars, we will calculate the variation of brightness between the planet Neptune and the planet Venus. There are three necessary factors: average brightness of each planet, ratio of brightness and the variation of brightness.<br />
<br />
=====Average Brightness=====<br />
Neptune: 7.9<br />
Venus: -4.355<br />
<br />
=====Ratio of brightness=====<br />
<p>x = mf - mb</p><br />
<p>Mf = magnitude of the fainter planet</p><br />
<p>Mb = magnitude of the brighter planet</p><br />
<p>http://i.imgur.com/amuUv.png</p><br />
<p> the ratio of the brightness scale can be seen above as I,II,III,IV, and V</p><br />
<br />
=====Variation of brightness=====<br />
<p><math>Vb = 2.512^x</math></p><br />
<br />
<p>To be able to find the variation in brightness, you first need to find the ''difference in magnitude'' by using the equation for the ratio of brightness.</p><br />
<p>x = mf - mb</p><br />
<p>x = 7.9 - -4.355</p><br />
<p>x = 12.255</p><br />
<br />
<p> Then, plug in the the ratio of brightness (x = 12.255) into the variation of brightness. </p><br />
<p><math>Vb = 2.512^x</math></p><br />
<p><math>Vb = 2.512^{12.255}</math></p><br />
<p>Vb = 79843.695</p><br />
<br />
<p>Therefore, Venus is 7843.695 times brighter than Neptune.</p><br />
<br />
===!!!HEY TEAM SEND ME THE PIC YOU WANT ME TO CROP ONTO THE STAR!!!===<br />
MY EMAIL IS magasparian@gmail.com</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_13&diff=74823Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 132011-02-03T06:41:07Z<p>Narissarasu: </p>
<hr />
<div>===Brightness of Stars on a logarithmic scale===<br />
<br />
When observing stars through the naked eye we notice that some are brighter than others. But is it possible to determine how bright a star is based upon a mathematical denotation. Lets read on... :)<br />
<br />
When determining star brightness, we must find a scale into which compare them to. Stars are placed on a 1-6 magnitude scale in terms of brightness (1 being the brightest, and 6 being the least brightest). The difference in magnitude from 1 to 2 is <math> \sqrt[5]{100}=2.512</math>. The reason why we use this number is because the difference in terms of brightness from 1-6 magnitude 1 is 100x greater than in 6<br />
<br />
In order to continue our explanation the logarithmic scale of the brightness of stars, we will calculate the variation of brightness between the planet Neptune and the planet Venus. There are three necessary factors: average brightness of each planet, ratio of brightness and the variation of brightness.<br />
<br />
=====Average Brightness=====<br />
Neptune: 7.9<br />
Venus: -4.355<br />
<br />
=====Ratio of brightness=====<br />
<p>x = mf - mb</p><br />
<p>Mf = magnitude of the fainter planet</p><br />
<p>Mb = magnitude of the brighter planet</p><br />
<p>http://i.imgur.com/amuUv.png</p><br />
<br />
=====Variation of brightness=====<br />
<p><math>Vb = 2.512^x</math></p><br />
<br />
<p>To be able to find the variation in brightness, you first need to find the ''difference in magnitude'' by using the equation for the ratio of brightness.</p><br />
<p>x = mf - mb</p><br />
<p>x = 7.9 - -4.355</p><br />
<p>x = 12.255</p><br />
<br />
<p> Then, plug in the the ratio of brightness (x = 12.255) into the variation of brightness. </p><br />
<p><math>Vb = 2.512^x</math></p><br />
<p><math>Vb = 2.512^{12.255}</math></p><br />
<p>Vb = 79843.695</p><br />
<br />
<p>Therefore, Venus is 7843.695 times brighter than Neptune.</p><br />
<br />
===!!!HEY TEAM SEND ME THE PIC YOU WANT ME TO CROP ONTO THE STAR!!!===<br />
MY EMAIL IS magasparian@gmail.com</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73217Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T23:10:04Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
Solution can be found in terms of t which represents the carrying capacity<br />
</p><br />
<br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif<br />
<br />
<br />
<br />
<p><br />
<br />
* economics side note ( possibly bonus marks for this??? :) )<br />
In actuality however, when examining the exponential growth of a particular population, let it be noted that they will never reach a specific asymptote. This occurs because an individuals offspring(from that population) will offset the asymptote constantly fluctuating above and below it. <br />
</p><br />
<br />
[[Image:EconpovScreen_shot_2011-01-27_at_3.05.59_PM.png]]<br />
</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73205Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T23:03:56Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
Solution can be found in terms of t which represents the carrying capacity<br />
</p><br />
<br />
<br />
<br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73203Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T23:01:23Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
<math> 50=\frac{95}{1 + e^{-t}} \!</math><br />
</p><br />
<p><br />
</math> ln{-45}=\ln{95}e^{-t}} \!</math><br />
</p><br />
<p><br />
<br />
</math> t=-1\ln{-45}/ln{95e}} \!</math><br />
</p><br />
<br />
<p><br />
Solution in terms of t represents the carrying capacity<br />
</p><br />
<br />
<br />
<br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73197Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:55:15Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
Solution in terms of t represents the carrying capacity<br />
</p><br />
<br />
<br />
<br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73192Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:54:03Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
Solution in terms of t represents the carrying capacity<br />
</p><br />
<br />
<br />
<br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73189Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:53:32Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
Solution in terms of t represents the carrying capacity<br />
</p><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p><br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73186Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:48:54Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<p><br />
<math>100x95% = 95 </math> polar bears<br />
</p> <br />
<p><br />
P(t) = 95<br />
</p><br />
<p><br />
<br />
<!-- :<math>95 = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>95 = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p><br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73182Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:41:58Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<p><br />
P = Population of polar bears <br />
</p><br />
<p><br />
t = time <br />
</p><br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p><br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73180Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:39:09Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 50 polar bears, however since it's carrying capacity is 100 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{50}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{50}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p><br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73179Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:38:31Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<p><b><br />
Example Model <br />
</p></b><br />
<br />
<p><br />
A small region in Antarctica holds a population of 100 polar bears, however since it's carrying capacity is 150 polar bears. Assume that the population of these polar bears are able to reproduce, by using the logistic growth model one is able to determine the exponential growth which will be obtain by the polar bear population. With the logistic growth model (displayed below) we intend to predict when the population reaches 95% of its carrying capacity. <br />
</p><br />
<br />
<br />
<p><br />
<br />
<!-- :<math>P(t) = \frac{100}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{100}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p><br />
<br />
http://johnbolton.ca/wp-content/uploads/2010/09/PolarBearParty1.gif</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73170Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:19:25Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73169Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:18:57Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<p><b><br />
Diagram 1<br />
</p></b><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> <br />
'''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> <br />
'''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+2}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/308wR.png<br />
</p><br />
<br />
<p> <br />
'''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+0.5}} \! </math><br />
</p><br />
<br />
<p><br />
http://i.imgur.com/KXQW7.png<br />
</p><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73165Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T22:12:22Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<p><b><br />
Diagram 1<br />
</p></b><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "K" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{K + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{K + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "K" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t+K}} \! </math><br />
</p><br />
<br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=73103User:Narissarasu2011-01-27T16:26:12Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally understand Calculus soon. <br />
<br />
<br />
<br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently taking both business as well as some economics courses, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Consumers and <br />
businesses face decisions such as whether or not to input extra hours at the workplace, buy a new equipment, or even to build an additional production facility daily. That is when a common economic tool known as the marginal analysis can help the business people make their decisions more productively and more efficiently, marginal analysis can also help these business people with making not only the right decision but also the decisions in which they would achieve the greatest benefit when apply, therefore it is very important for both business people and economist to know how to obtain the marginal cost. First of all, what is Marginal cost? marginal cost is the increase or decrease in the total costs of a business company as a result of one more or one less unit of output produced or added into the firm. Determining marginal cost is important in deciding whether or not to vary a rate of production, in another word it is defined as the change in total cost that arises when one extra unit of product is being produced. A sale at a price higher than marginal unit cost will increase the net profit of the manufacturer even though the sales price does not cover average total unit cost; marginal cost is then the lowest amount at which a sale can be made without adding to the loss of producers or subtracting from his or her profits. (The marginal cost diagram can be seen below).<br />
<p> <br />
http://i.imgur.com/zKVDd.png<br />
</p><br />
<br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will benefit from taking the action. <br />
Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/jMIYw.png<br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=73102User:Narissarasu2011-01-27T16:24:55Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally understand Calculus soon. <br />
<br />
<br />
<br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently taking both business as well as some economics courses, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Consumers and <br />
businesses face decisions such as whether or not to input extra hours at the workplace, buy a new equipment, or even to build an additional production facility daily. That is when a common economic tool known as the marginal analysis can help the business people make their decisions more productively and more efficiently, marginal analysis can also help these business people with making not only the right decision but also the decisions in which they would achieve the greatest benefit when apply, therefore it is very important for both business people and economist to know how to obtain the marginal cost. First of all, what is Marginal cost? marginal cost is the increase or decrease in the total costs of a business company as a result of one more or one less unit of output produced or added into the firm. Determining marginal cost is important in deciding whether or not to vary a rate of production, in another word it is defined as the change in total cost that arises when one extra unit of product is being produced. A sale at a price higher than marginal unit cost will increase the net profit of the manufacturer even though the sales price does not cover average total unit cost; marginal cost is then the lowest amount at which a sale can be made without adding to the loss of producers or subtracting from his or her profits. (The marginal cost diagram can be seen below).<br />
<p> <br />
http://i.imgur.com/jEQoc.png<br />
</p><br />
<br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will benefit from taking the action. <br />
Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/t4OH1.png <br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=73101User:Narissarasu2011-01-27T16:21:02Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally understand Calculus soon. <br />
<br />
<br />
<br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently taking both business as well as some economics courses, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Consumers and <br />
businesses face decisions such as whether or not to input extra hours at the workplace, buy a new equipment, or even to build an additional production facility daily. That is when a common economic tool known as the marginal analysis can help the business people make their decisions more productively and more efficiently, marginal analysis can also help these business people with making not only the right decision but also the decisions in which they would achieve the greatest benefit when apply, therefore it is very important for both business people and economist to know how to obtain the marginal cost. First of all, what is Marginal cost? marginal cost is the increase or decrease in the total costs of a business company as a result of one more or one less unit of output produced or added into the firm. Determining marginal cost is important in deciding whether or not to vary a rate of production, in another word it is defined as the change in total cost that arises when one extra unit of product is being produced. A sale at a price higher than marginal unit cost will increase the net profit of the manufacturer even though the sales price does not cover average total unit cost; marginal cost is then the lowest amount at which a sale can be made without adding to the loss of producers or subtracting from his or her profits. (The marginal cost diagram can be seen below).<br />
<p> <br />
http://i.imgur.com/jEQoc.png <br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will benefit from taking the action. <br />
Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/t4OH1.png <br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73089Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T07:27:34Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<p><b><br />
Diagram 1<br />
</p></b><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><b><br />
Function: <br />
</p></b><br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
Graph:<br />
</p><br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "1" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<p><b><br />
graph:<br />
</p></b><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<p><b><br />
Diagram 2<br />
</p></b><br />
<br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<p><b>Diagram 3</p></b><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73088Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T07:26:52Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<p><b><br />
Diagram 1<br />
</p></b><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
Function: <br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
Graph:<br />
</p><br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "1" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
<p><b><br />
Function:<br />
</p></b><br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<p><b><br />
graph:<br />
</p></b><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<p><b><br />
Diagram 2<br />
</p></b><br />
<br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<p><b>Diagram 3</p></b><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73085Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T07:23:33Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<p><b><br />
Diagram 1<br />
</p></b><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
Function: <br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
Graph:<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "1" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
function:<br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
<p><b><br />
graph:<br />
</p></b><br />
[[Image:NewlowerMSP279819e48gb1e186b0ia00002578hi9a0dfd949d.gif]]<br />
<p><b><br />
Diagram 2<br />
</p></b><br />
<br />
[[Image:NewhigherMSP1032619e47h3ahi5g723f0000296bihh7agg22a26.gif]]<br />
<p><b>Diagram 3</p></b><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73075Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T07:17:42Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
Function: <br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
Graph:<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "1" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
function:<br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
graph:<br />
<br />
<small>[[Image:LowerinterceptScreen_shot_2011-01-26_at_10.59.02_PM.png]]</small><br />
<br />
<small>[[Image:HigherinterceptScreen_shot_2011-01-26_at_10.56.51_PM.png]]</small><br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73072Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T07:16:17Z<p>Narissarasu: </p>
<hr />
<div><p> '''original formula'''<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<p><br />
[[Image:MiddleinterceptMSP1356819e47a6f1ff534dd00001d4hf0i8dief886h.gif]]<br />
</p> <br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
Function: <br />
<br />
<p> '''wider asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p>'''narrower asymptote'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
Graph:<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
definition: Similarly to the question above, in order to manipulate the y-intercept to any other number between 0 and K you must also change the "1" in the denominator of the original formula. By inputting a number larger than 1, the y-intercept crosses on a lower intercept than in the original formula. On the other hand, if you input a smaller number in the denominator rather than 1, you may observe that the y-intercept is at a higher point on the graph.<br />
<br />
function:<br />
<br />
<p> '''lower intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{2 + e^{-t}} \! </math><br />
</p><br />
<br />
<p> '''higher intercept'''<br />
<br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{.5 + e^{-t}} \! </math><br />
</p><br />
graph:<br />
<br />
[[Image:LowerinterceptScreen_shot_2011-01-26_at_10.59.02_PM.png|center]]<br />
<br />
[[Image:HigherinterceptScreen_shot_2011-01-26_at_10.56.51_PM.png|center]]<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=73038Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-27T06:12:40Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
<br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72963Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T23:08:31Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
<p><i><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><i><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72957Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T23:05:41Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
<p><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72956Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T23:05:25Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<br />
<br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
<p><br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><br />
<br />
<br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72953Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T23:04:08Z<p>Narissarasu: </p>
<hr />
<div><br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<br />
<p><br />
<br />
In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
</p><br />
<br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72951Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T23:02:43Z<p>Narissarasu: </p>
<hr />
<div>In order to manipulate the height of the horizontal asymptote, you must change "1" in the denominator to a larger number if you wish for a broader asymptote and to a smaller number if you wish for a closer asymptote<br />
<br />
* Change the y-intercept to any number between 0 and K<br />
<br />
<p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p><br />
<br />
<br />
<p><br />
- Change the height of the horizontal asymptote on the right and denote it by K.<br />
</p><br />
<p><br />
- Change the y-intercept to any number between 0 and K.<br />
</p><br />
<p><br />
- BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.<br />
</p><br />
<br />
<br />
<p><b><br />
Solution<br />
</p></b><br />
<br />
<p><br />
<math>P(t)=\frac{K}{1+e^{-t}}\quad</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72924Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T22:34:28Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<!-- :<math>P(t) = \frac{1}{1 + e^{-t}} \! </math> --> <br />
:<math>P(t) = \frac{1}{1 + e^{-t}} \! </math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72919Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T22:28:12Z<p>Narissarasu: </p>
<hr />
<div><p><br />
<math>P(t)=1/(1+(e^(-t))</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework_12&diff=72917Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework 122011-01-26T22:27:45Z<p>Narissarasu: Created page with "<p> <math>P(t)=(1/(1+e^(-t))</math> </p>"</p>
<hr />
<div><p><br />
<math>P(t)=(1/(1+e^(-t))</math><br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=72296User:Narissarasu2011-01-25T22:16:43Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally understand Calculus soon. <br />
<br />
<br />
<br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently studying both business as well as economics, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Consumers and <br />
businesses face questions such as whether to put in a few extra hours at work, save a little extra each month, buy a new computer, or build an additional production facility everyday. A central tool of economic research known as marginal analysis can provide decision makers with the tools for making decisions that will achieve the greatest benefit, and this is when Marginal cost analysis becomes very important. So what is marginal cost? Marginal cost is the increase or decrease in the total costs of a business firm as the result of one more or one less unit of output. Determining marginal cost is important in deciding whether or not to vary a rate of production in another word it is defined as the change in total cost that arises when one extra unit of product is being produced. A sale at a price higher than marginal unit cost will increase the net profit of the manufacturer even though the sales price does not cover average total unit cost; marginal cost is thus the lowest amount at which a sale can be made without adding to the producer's loss or subtracting from his profits.<br />
<p> <br />
http://i.imgur.com/jEQoc.png <br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will be making profits. Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/t4OH1.png <br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=72292User:Narissarasu2011-01-25T22:16:28Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally understand Calculus soon. <br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently studying both business as well as economics, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Consumers and <br />
businesses face questions such as whether to put in a few extra hours at work, save a little extra each month, buy a new computer, or build an additional production facility everyday. A central tool of economic research known as marginal analysis can provide decision makers with the tools for making decisions that will achieve the greatest benefit, and this is when Marginal cost analysis becomes very important. So what is marginal cost? Marginal cost is the increase or decrease in the total costs of a business firm as the result of one more or one less unit of output. Determining marginal cost is important in deciding whether or not to vary a rate of production in another word it is defined as the change in total cost that arises when one extra unit of product is being produced. A sale at a price higher than marginal unit cost will increase the net profit of the manufacturer even though the sales price does not cover average total unit cost; marginal cost is thus the lowest amount at which a sale can be made without adding to the producer's loss or subtracting from his profits.<br />
<p> <br />
http://i.imgur.com/jEQoc.png <br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will be making profits. Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/t4OH1.png <br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=72226User:Narissarasu2011-01-25T21:50:57Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really a big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally "understand" Calculus soon! <br />
<br />
<p><b> <br />
Homework 12 Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you. <br />
</p></b> <br />
<p> <br />
I'm currently studying both business as well as economics, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Average and Marginal Cost for example, marginal cost is defined as the change in total cost that arises when one extra unit of product is being produced. <br />
<p> <br />
http://i.imgur.com/jEQoc.png <br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will be making profits. Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p> <br />
http://i.imgur.com/t4OH1.png <br />
</p> <br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=72027User:Narissarasu2011-01-25T01:32:11Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really a big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally "understand" Calculus soon!<br />
<br />
<br />
<p><b><br />
Homework 12 <br />
Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you.<br />
</p></b><br />
<p><br />
I'm currently studying both business as well as economics, therefore I thought I might just as well talk about how calculus can be apply in the business decision making since economics, business, and calculus are closely related to each other. Average and Marginal Cost for example, marginal cost is defined as the change in total cost that arises when one extra unit of product is being produced. <br />
<p><br />
http://i.imgur.com/jEQoc.png<br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will be making profits.<br />
Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p><br />
http://i.imgur.com/t4OH1.png<br />
</p><br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=71716User:Narissarasu2011-01-24T02:14:54Z<p>Narissarasu: </p>
<hr />
<div>Hi, I'm Narissa, first year student at UBC (Sauder School of Business). I enjoy learning math but never really a big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally "understand" Calculus soon!<br />
<br />
<br />
<p><b><br />
Homework 12 <br />
Write an essay describing a particular use of calculus in your field of study, or in a field of interest to you.<br />
</p></b><br />
<p><br />
I'm currently studying both business as well as economics, therefore I thought I might just as well talk about how calculus is used in the business decision making since economics, business, and calculus are closely related. Average and Marginal Cost for example, marginal cost is the change in total cost that arises when one extra unit of product is being produced. <br />
<p><br />
http://i.imgur.com/jEQoc.png<br />
</p><br />
In order to figure out the marginal cost of producing one extra unit of output or what is the marginal cost of hiring one extra unit of employee or what is the marginal benefit of opening an extra unit of company? Business people as well as economist are often time more concern with the cost arise from the extra unit of the product or output, since this would provide them with a more accurate detail analysis of whether the company as a whole will be making profits.<br />
Marginal cost is derived from the derivative of the total cost. For example, given the total cost equation of <math>C(x) = -0.02(x^2) + 50x +100</math> as seen in the below diagram. <br />
<p><br />
http://i.imgur.com/t4OH1.png<br />
</p><br />
To find the marginal cost for example at 100 items, we need to do a derivative of <math>C(x)</math>, the result <math>C'(x) = 46</math> we can conclude that the marginal cost for producing the 101th unit of output will cost $46 dollars. When the marginal benefit of the production is higher than the marginal cost this will then mean it would cost less for the company to produce, whereas if the marginal benefit is lower than the marginal cost, this means that the company will be making a loss of producing each extra unit of output since the firm will be paying higher than they receive. The decision for the business people will then be, the company will only produce if and only if the marginal benefit is greater or equal to the marginal cost of the production. <br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70580Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T23:20:15Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<p><br />
C = cost of the production<br />
</p><br />
<p><br />
X = the quantity of the production<br />
</p><br />
<p><br />
after the production of the initial 20 units (with a total cost of 100$), the cost of producing an additional unit is $7 each. Therefore the marginal cost of producing an additional unit increases. Moreover, if the cost of producing the first 20 units were 7$ each, the total cost of producing te first 20 units would be 140$ rather than the 100$. This implies a 40$ change from the original cost of production which is then equated into our formula <math>C = 7X - 40</math>.<br />
</p><br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/OfSBu.png<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
A average fixed cost is calculated by dividing total cost by the quantity produced ,the graph is represented graphically as an ever decreasing asymptotic to the horizontal axis. An example being, the rent paid by a restaurant is divided among more and more meals as the volume of production increses, the average cost of per meals attributable to the fixed rent decreases as the number of meals increase. General formula: P=1/Q as X>0<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as Y = ((<math>X^2</math>) – 20) + 30.<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70570Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T23:07:10Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/OfSBu.png<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
A average fixed cost is calculated by dividing total cost by the quantity produced ,the graph is represented graphically as an ever decreasing asymptotic to the horizontal axis. An example being, the rent paid by a restaurant is divided among more and more meals as the volume of production increses, the average cost of per meals attributable to the fixed rent decreases as the number of meals increase. General formula: P=1/Q as X>0<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as Y = ((<math>X^2</math>) – 20) + 30.<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70569Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T23:06:31Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/OfSBu.png<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
A average fixed cost is calculated by dividing total cost by the quantity produced ,the graph is represented graphically as an ever decreasing asymptotic to the horizontal axis. An example being, the rent paid by a restaurant is divided among more and more meals as the volume of production increses, the average cost of per meals attributable to the fixed rent decreases as the number of meals increase. General formula: P=1/Q as X>0<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as Y = ((<math>X^2</math>) – 20) + 30.<br />
</p><br />
<br />
<br />
<br />
<br />
<p><br />
Matthew is awesome ^.^<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70423Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T15:56:37Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/53gdF.jpg<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as <math> Y = ((X^2) – 20) + 30 </math>.<br />
</p><br />
<br />
<br />
<br />
<br />
<p><br />
Matthew is awesome ^.^<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70422Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T15:56:10Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/53gdF.jpg<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as <math> Y = ((X^2) – 20) + 30 </math>.<br />
</p><br />
<br />
<br />
<br />
<br />
<p><br />
Matthew is awesome ^^<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70421Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T15:53:58Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/53gdF.jpg<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as <math> Y = ((X^2) – 20) + 30 </math>.<br />
</p><br />
<br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<p><br />
Matthew is awesome ^^<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70420Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T15:51:43Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/53gdF.jpg<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/D6RLg.png<br />
</p><br />
<br />
<br />
<p> <br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
We can look at the graph of this function to see the relationship between the average cost and production. <br />
</p><br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as <math> Y = ((X^2) – 20) + 30 </math>.<br />
</p><br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<p><br />
=^.^=<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Teams/St_Gallen/Homework/11_Part3&diff=70419Course:MATH110/Archive/2010-2011/003/Teams/St Gallen/Homework/11 Part32011-01-18T15:50:33Z<p>Narissarasu: </p>
<hr />
<div><p><b><i><br />
Write a linear model to predict the cost of producing flags of your team's Canton under the assumptions that the marginal cost is $7 per unit and that at the current production level of 20 items, the cost is $100.<br />
</b></i></p><br />
<br />
<p><br />
The linear model that we came up with was <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Describe your model.<br />
</b></i></p><br />
<br />
<p><br />
<math>C = 7X - 40</math><br />
</p><br />
<br />
<br />
<br />
<p><b><i><br />
What does your model predict for a production of 150 items?<br />
</b></i></p><br />
<br />
<p><br />
By using the above equation <math>C = 7X - 40</math>. Where C = cost of the production in dollars; X = the amount of items being produced. We substitute 150 items into the X to find the cost of the total production, <math>C = 7(150) - 40; C = 1010 </math>. <br />
</p><br />
<br />
<br />
<p><b><i><br />
According to your model, what happens to the average cost per item as production levels increase?<br />
</b></i></p><br />
<br />
<p> <br />
According to our model, the average cost per item, as production levels will increase by approximately 1.73 dollars per each extra item produced. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost remains constant as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/53gdF.jpg<br />
</p><br />
<br />
<p><br />
The diagram above shows a constant average cost, as extra product is being produce. As given the quantity of 20 items, the price for the output remains the same and it is given the cost of $100 for the production of 20 items. Therefore the model for the above can be given as Y = 100. <br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost diminishes as production increases.<br />
</b></i></p><br />
<br />
<p><br />
<br />
</p><br />
<br />
<br />
<p><b><i><br />
The average cost increases as production increases.<br />
</b></i></p><br />
<br />
<p><br />
http://imgur.com/D6RLg<br />
</p><br />
<br />
<br />
<p><br />
<br />
The model P=Q+2^Q where P is price and Q is the number of items increases the average cost exponentially as production increases.<br />
<br />
We can look at the graph of this function to see the relationship between the average cost and production.<br />
<br />
</p><br />
<br />
<p><b><i><br />
You obtain an economy of scale. This means that starting at some specific production level, the marginal cost is always less than the average cost.<br />
</b></i></p><br />
<br />
<p><br />
http://i.imgur.com/RCyfa.jpg<br />
</p><br />
<br />
<p><br />
In the above diagram, there is an increase in the average cost as extra unit of production is being produced, this is also known as the reverse of economic of scales. As the production increases from 25 items to 30 items, we can see that there is an increase in the cost of producing the product from $60 to $110. Therefore the average cost for producing 25 items will increase approximately 20% in the price. The model for the above diagram could be written as <math> Y = ((X^2) – 20) + 30 </math>.<br />
</p><br />
<br />
<p><br />
As you can see from the graph as the quantity increases the average cost increases. 3 items will cost $11 while 8 items will cost $264. The average cost for the 3 items is $3.66 while the average cost for items is $33. <br />
</p><br />
<br />
<br />
<p><b><i><br />
Other interesting properties that you can think of and create a model for. <br />
</b></i></p><br />
<br />
<p><br />
=^.^=<br />
</p></div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Webwork_Week_8&diff=58559Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork Week 82010-10-29T21:48:20Z<p>Narissarasu: </p>
<hr />
<div>==Problem 8==<br />
I am having difficulty getting this one:<br />
<br />
h(x) = <math> sqrt{(9x/x^2-361)} </math> <br />
<br />
Given that -19 and 19 for x would cause the denominator to equal 0, I thought this was the answer:<br />
<br />
(-infinity,-19)U(-19,19)U(19,infinity)<br />
<br />
but that is incorrect. Maybe I am thinking about this in the wrong way. Anyways, I am stuck on this one and would appreciate a hint. <br />
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 20:05, 28 October 2010 (UTC)<br />
<br />
<br />
<br />
I also have the same values as you, and what I have seen is that -19 and 19 don't work, but numbers between 1 through 18 don't work either because the bottom will be a negative number meaning the overall thing will be negative. So I have tried (-infinity, -19)U(-19,0]U(19,infinity) but that is also incorrect. I think it might be something about me excluding numbers between 1 through 18. I still haven't found the answer, but if that helps you find it.<br />
--[[User:JoseTorresTorija|JoseTorresTorija]]<br />
<br />
<br />
Hey Steffany <br />
<p>So you were given that -19 and 19 would make the denominator equal to 0, recall that we could not have any negative value in the square root as it would be undefined. <br />
<p>Therefore I got (-19,0]U(19,infinity), try this and I think it should work. <br />
--[[User:Narissarasu|Narissarasu]]<br />
<br />
==Q6==<br />
Was Question 6 removed from the Webwork?<br />
<br />
I think so, I realized one of the questions was missing.<br />
<br />
Because it had a bug -- [[User:DavidKohler|DavidKohler]]<br />
==Problem 4==<br />
Can anyone help me out with question 4?<br />
<br />
Problem 4 is tricky. Here is my hint for solving it. Think of each job as a different option. So you have two options which can be represented by two different formulas (I will let you work out the formulas). The key to answering this is the wording "at least as good." Another way of saying at least as good is saying "equal." Once you have your two formulas, think of what you do with the formulas to solve this...<br />
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 20:05, 28 October 2010 (UTC)</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=Course:MATH110/Archive/2010-2011/003/Math_Forum/Lecture18&diff=57477Course:MATH110/Archive/2010-2011/003/Math Forum/Lecture182010-10-24T00:30:36Z<p>Narissarasu: </p>
<hr />
<div>Thanks for posting Bernadette! but I think some of your code got a little mixed up when you posted...<br />
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 21:05, 23 October 2010 (UTC)<br />
----<br />
Hi Steffany,<br />
So I've worked out your questions.<br />
<br />
The limit as X approaches infinity of <math>sqrt(X^2 + 13x - 2)</math>- x <br />
<br />
Think of this as <math>sqrt((X^2+13x-2)-x)/1)</math>*<math>((sqrt(X^2+13x-2)+x)/(sqrt(X^2+13x-2)+x))</math><br />
<br />
What this step does is it rationalizes the numerator to simplify the equation.<br />
<br />
You should get<br />
<br />
<math>(x^2+13x-2-x^2)/(sqrt(X^2+13x-2)+x)</math><br />
<br />
Divide the numerator and denominator by the biggest X power found in the denominator, which in this case is X.<br />
<br />
You should get <br />
<math>(X+13-(2/x)-x)/(1+(sqrt13x/x)-(sqrt2/x)+1)</math><br />
<br />
which simplifies to <br />
<math>13/2</math><br />
<br />
It's the same types of steps for the second problem, try that,and if it still doesn't work let me know! --[[User:BernadetteHii|BernadetteHii]] 18:46, 23 October 2010 (UTC)<br />
<br />
<br />
I talked to a couple of people in class today who had trouble with this one from the Lecture 18 set yesterday, so I think it is worth mentioning. And yes, I'm on the math forum on a Friday night, so I don't expect to hear back from anyone right away...<br />
<br />
The last 2 parts of question 5:<br />
<br />
The limit as X approaches infinity of <math>sqrt(X^2 + 13x - 2)</math>- x and <math>sqrt(X^2 + 13x - 2)</math> + x<br />
<br />
Any hints to solve these ones would be helpful. I tried several different ways, like pulling out x^2 first, but every time I solve it I get negative infinity for the first one. I haven't had problems with the rest of the ones involving square roots when solving them this way so I'm stuck. <br />
<br />
Victoria - about 4(a) - although you may have asked around today and figured this out already - Look at what you are left with after you have pulled the x out of numerator and the x^2 out of the denominator. My hint is to remember that when you square a number x and then take the root of it you are left with...<br />
--[[User:SteffanyChwedoruk|SteffanyChwedoruk]] 01:04, 23 October 2010 (UTC)<br />
----<br />
Has anyone got any suggestions for #4a)? I've come up with so many different answers and none of them are right. Any helpful hints out there? (and is anyone else totally frustrated and feeling completely lost on these?) ANY advice on any of the asymptotes is helpful. I felt like the Wiki notes were helpful for the concept but not for the computations. <br />
Thanks- Victoria<br />
<br />
<br />
<br />
Hi Victoria<br />
4a) is similar to the previous questions, so what you have to do first is to sqare-root the denominator then divide both the numerator and the denominator by the largest t power found in the denominator. <br />
<br />
My #4a) on the webwork was:<br />
The limit as t approaches infinity <math>(-1t - 9)/(sqrt(t^2 + 6t + 5)</math>)<br />
<br />
what I did first was I simplify the denominator <math>(sqrt(t^2 + 6t + 5)</math>)<br />
and I got <math>(t + sqrt(6t) + sqrt(5)) </math><br />
<br />
after that I divide both the denominator <math>(t + sqrt(6t) + sqrt(5)) </math> and the numerator <math>(-1t - 9)</math> by the largest t power, in this case it would be <math>(t^1)</math> <br />
you will then get <math>((-1t - 9)/t)/(t + (sqrt(6t)/t) + (sqrt(5)/t))</math> but since t cancels out we now left with <math>(-1 - (9/t))/(1 + (sqrt(6t)/t) + (sqrt(5)/t))</math><br />
therefore as the limit t approaches infinity we should get -1.<br />
<br />
Let me know if you need a more detail explanation on this question <br />
--[[User:Narissarasu|Narissarasu]] 17:32, 23 October 2010 (UTC)<br />
----</div>Narissarasuhttps://wiki.ubc.ca/index.php?title=User:Narissarasu&diff=47978User:Narissarasu2010-09-19T17:07:10Z<p>Narissarasu: Created page with 'Hi, I'm Narissa a first year student at UBC (Sauder School of Business). I enjoy learning math but never really a big fan of Calculus and Differentiation, but I'm excited to be i…'</p>
<hr />
<div>Hi, I'm Narissa a first year student at UBC (Sauder School of Business). I enjoy learning math but never really a big fan of Calculus and Differentiation, but I'm excited to be in math 110 and I'm looking forward to finally "understand" Calculus soon!</div>Narissarasu