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Science:Math Exam Resources

2014-09-26T01:08:32Z

CindyBlois:

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<div style="font-size:170%; line-height:100%; border:none; margin:0; padding:0em; color:#FFFFFF; font-weight:bold; line-height:1.3;">Welcome to the <br> Math Exam / Education Resources wiki</div>
<div style="padding-left:+0.5em; margin-top:+0.5em; font-size:140%; line-height:100%; border-left:2px solid #FFFFFF; line-height:1.3;"><i>a place to learn about mathematics <br> created by the UBC mathematical community</i></div>
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<li>Fully solved questions: <b>{{MER count QGS}}</b></li>
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Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 04/Solution 1

2014-04-11T01:45:32Z

CindyBlois:

__NOTOC__
We reduce the matrix to row-echelon form. Subtracting the first row from the last row transforms

:<math>
M = \begin{bmatrix} 1 & 7 & 3 \\ 0 & p & 5 \\ 1 & pq + 7 & 8 \end{bmatrix}
</math>

into

:<math>
\begin{bmatrix} 1 & 7 & 3 \\ 0 & p & 5 \\ 0 & pq & 5 \end{bmatrix}.
</math>

Now, subtracting the second row from the third row yields

:<math>
\begin{bmatrix} 1 & 7 & 3 \\ 0 & p & 5 \\ 0 & pq-p & 0 \end{bmatrix}
</math>


This matrix either has rank 2 or rank 3. If ''pq - p=p(q - 1) = 0'', then ''M'' has rank 2; otherwise, ''M'' has rank 3. In other words, ''M'' has rank 2 exactly when ''p=0'' or ''q=1''.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 04

2014-04-11T01:36:50Z

CindyBlois:

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[[Category:MER Tag Rank and nullity]]

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Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 04/Hint 1

2014-04-11T01:36:03Z

CindyBlois:

What does it mean for the matrix to have rank 2? In row-echelon form, the first two rows must have non-zero entries, while the last row must consist of zeros.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 04/Hint 1

2014-04-11T01:32:35Z

CindyBlois:

Perform row operations. As a start, the first and last rows look very similar.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 03

2014-04-11T01:29:48Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 03/Solution 1

2014-04-11T01:28:50Z

CindyBlois:

<math>a </math> and <math>d </math> are orthogonal whenever their dot product is equal to zero. That is,

:<math>
\begin{align}
0 = a \cdot d &= \begin{bmatrix}
-3 \\ -3 \\ 3
\end{bmatrix}
\cdot
\begin{bmatrix} -1 \\ k \\ 2k+1 \end{bmatrix}\\
&= (-3)(-1) + (-3)(k) + (3)(2k+1)\\
&= 3 - 3k + 6k+3\\
&= 6 + 3k\\
&= 3(2 + k).
\end{align}
</math>

Therefore, <math> \displaystyle k = -2 </math> is the value for which <math>a </math> and <math>d </math> are orthogonal.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02

2014-04-11T01:20:23Z

CindyBlois:

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[[Category:MER Tag Projections]]

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Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Solution 1

2014-04-11T01:19:51Z

CindyBlois:

We will use the formula,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c .\end{align}</math>

Note that

:<math>a\cdot c = -9+12+36 = 39</math>

and

:<math>|c|^2 = c\cdot c = 9+16+144 = 169.</math>

Hence,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c = \frac{39}{169} c = \frac{3}{13}\begin{bmatrix}
3 \\ -4 \\ 12
\end{bmatrix} .\end{align}</math>

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Solution 1

2014-04-11T01:17:45Z

CindyBlois:

We will use the formula,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c .\end{align}</math>

Note that

:<math>a\cdot c = -9+12+36 = 39</math>

and

:<math>|c|^2 = c\cdot c = 9+16+144 = 169.</math>

Therefore,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c = \frac{39}{169} c = \frac{3}{13}\begin{bmatrix}
3 \\ -4 \\ 12
\end{bmatrix} .\end{align}</math>

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Solution 1

2014-04-11T01:16:52Z

CindyBlois: Created page with " We will use the formula, :<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c .\end{align}</math> Note that :<math>a\cdot c = -9+12+36 = 39</math> and :<ma..."

We will use the formula,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c .\end{align}</math>

Note that

:<math>a\cdot c = -9+12+36 = 39</math>

and

:<math>|c|^2 = c\cdot c = 9+16+144 = 169.</math>

Therefore,

:<math>\begin{align} \text{Proj}_{c}a &= \frac{a\cdot c}{|c|^2} c = \frac{39}{169} c = \frac{3}{13} .\end{align}</math>

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Hint 2

2014-04-11T01:06:59Z

CindyBlois:

:<math> \text{Proj}_{c}a = \frac{a\cdot c}{|c|^2} c </math>

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Hint 2

2014-04-11T01:06:25Z

CindyBlois: Created page with "<math> \text{Proj}_{c}a = \frac{a\cdot c}{|c|^2} c </math>"

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 02/Hint 1

2014-04-11T01:04:33Z

CindyBlois: Created page with "``<math> \text{Proj}_{c}a </math> " is the projection of <math> a</math> onto <math> c</math>. If you remember the formula for this, you may use it directly. If you don't re..."

``<math> \text{Proj}_{c}a </math> " is the projection of <math> a</math> onto <math> c</math>. If you remember the formula for this, you may use it directly. If you don't remember the formula, draw a diagram with <math> a</math>, <math> c</math>, and <math> \text{Proj}_{c}a </math>. Use the picture to help you deduce the formula.

Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 01

2014-04-11T00:51:39Z

CindyBlois:

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[[Category:MER Tag Matrix operations]]

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Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 01/Hint 2

2014-04-11T00:49:20Z

CindyBlois:

Use a cofactor expansion to find the determinant of a 3 by 3 matrix. Don't forget about the sign of each factor!

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)

2013-11-16T02:13:38Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)/Solution 1

2013-11-16T02:12:29Z

CindyBlois: Created page with "<math>f(1)=1 </math> (Explanation: We are given that the curve <math>x^2+y^3-2xy=0 </math> passes through the point <math> (1,1)</math>. Near this point, the curve is the ..."

<math>f(1)=1 </math>

(Explanation: We are given that the curve <math>x^2+y^3-2xy=0 </math> passes through the point <math> (1,1)</math>. Near this point, the curve is the graph of a function, <math> y=f(x) </math>. When <math>x=1 </math>, <math>y=f(x)=1 </math>.)

<math>f'(1) </math> is the same thing as <math> \frac{dy}{dx}</math> evaluated at the point <math>(x,y)=(1,1) </math>. In order to find this, we use implicit differentiate the equation <math>x^2+y^3-2xy=0 </math> with respect to <math>x </math>.

<math>
\begin{align}
2x+3y^2 \frac{dy}{dx}-2y-2x\frac{dy}{dx}=0
\end{align}
</math>

In order to solve for <math>\frac{dy}{dx} </math>, we first factorize,

<math>
\begin{align}
2x-2y+(3y^2-2x) \frac{dy}{dx}=0.
\end{align}
</math>

Then subtract <math>2x-2y</math> from both sides and divide by <math>(3y^2-2x)</math> to obtain

<math>
\begin{align}
\frac{dy}{dx}=\frac{2y-2x}{3y^2-2x}.
\end{align}
</math>

When <math>x=y=1,</math>

<math>
\begin{align}
\frac{dy}{dx}=\frac{2(1)-2(1)}{3(1)^2-2(1)}=0.
\end{align}
</math>

Hence, <math>f'(1)=0</math>.

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)/Hint 3

2013-11-16T01:45:24Z

CindyBlois: Created page with "Use implicit differentiation to find <math>\frac{d y}{dx} </math> and evaluate at <math>x=y=1 </math>."

Use implicit differentiation to find <math>\frac{d y}{dx} </math> and evaluate at <math>x=y=1 </math>.

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)/Hint 2

2013-11-16T01:44:14Z

CindyBlois: Created page with "Note that <math> f'(1)</math> is equal to <math>\frac{d y}{dx} </math> at the point <math>(x,y)=(1,1) </math>. How can you find <math>\frac{d y}{dx} </math>?"

Note that <math> f'(1)</math> is equal to <math>\frac{d y}{dx} </math> at the point <math>(x,y)=(1,1) </math>. How can you find <math>\frac{d y}{dx} </math>?

Science:Math Exam Resources/Courses/MATH104/December 2012/Question 06 (a)/Hint 1

2013-11-16T01:40:31Z

CindyBlois: Created page with "To find <math>f(1) </math>, note that the curve <math>y=f(x) </math> passes through the point <math>(1,1) </math>. When <math> x=1</math>, what is <math>f(x) </math>?"

To find <math>f(1) </math>, note that the curve <math>y=f(x) </math> passes through the point <math>(1,1) </math>. When <math> x=1</math>, what is <math>f(x) </math>?

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 5

2013-04-11T16:31:48Z

CindyBlois:

Note that we are trying to find the rate of change of the distance between the probe and ship, at the particular time of 10 s after the drop.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08

2013-04-10T22:18:11Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 5

2013-04-10T22:17:40Z

CindyBlois: Created page with "Note that we are trying to find <math>\frac{dr}{dt} </math>, at the particular time of 10 s after the drop."

Note that we are trying to find <math>\frac{dr}{dt} </math>, at the particular time of 10 s after the drop.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 4

2013-04-10T22:11:32Z

CindyBlois: Created page with "In order to find the relationship between the time derivatives of these variables, implicitly differentiate your equation with respect to time."

In order to find the relationship between the time derivatives of these variables, implicitly differentiate your equation with respect to time.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 3

2013-04-10T22:05:34Z

CindyBlois: Created page with "Use Pythagoras."

Use Pythagoras.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 2

2013-04-10T22:04:06Z

CindyBlois: Created page with "Can you find an equation that relates the distance between the probe and ship with the distances of the probe and ship from the drop point?"

Can you find an equation that relates the distance between the probe and ship with the distances of the probe and ship from the drop point?

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Hint 1

2013-04-10T22:01:22Z

CindyBlois: Created page with "Draw a picture."

Draw a picture.

Science:Math Exam Resources/Courses/MATH110/April 2012/Question 08/Solution 1

2013-04-10T22:00:49Z

CindyBlois: Created page with "Let <math> x</math> be the distance of the ship from the drop point, <math> y</math> be the distance of the probe from the drop point, and <math>r </math> be the distance betw..."

Let <math> x</math> be the distance of the ship from the drop point, <math> y</math> be the distance of the probe from the drop point, and <math>r </math> be the distance between the ship and the probe.

We want to find <math> \frac{dr}{dt}</math>, at time 10s after the probe is dropped. We know that <math> \frac{dx}{dt} = 5 m/s</math> and <math> \frac{dy}{dt} = 4 m/s</math>.

By the Pythagorean Theorem,
:<math> x^2 + y^2 = r^2.</math>

In order to find the relationship between the time rates of change of these variables, we differentiate both sides of this equation with respect to time, in order to obtain

:<math> 2x \frac{dx}{dt}+2y \frac{dy}{dt} = 2r\frac{dr}{dt}</math>

Dividing both sides by <math>2r </math> in order to isolate <math> \frac{dr}{dt}</math>, we find

:<math> \frac{dr}{dt} =\frac{x \frac{dx}{dt}+y \frac{dy}{dt}}{r}. </math>

We are interested in the value of <math> \frac{dr}{dt}</math>, at time <math>10s </math> after the probe is dropped. At this point in time,
:<math> x= 5 m/s \cdot 10s = 50 m,</math>
:<math> y= 4 m/s\cdot 10s = 40 m,</math>
and
:<math>r = \sqrt{x^2+y^2} = \sqrt{(50m)^2+(40m)^2} = 10\sqrt{5^2+4^2} \ m = 10\sqrt{41} \ m. </math>

Hence,
:<math> \frac{dr}{dt} = \frac{(50 m) (5 m/s)+(40 m)(4m/s) }{10\sqrt{41}m} =\frac{25m/s+16m/s}{\sqrt{41}} = \sqrt{41} \ m/s .</math>

That is, <math> 10s</math> after the probe is dropped, the distance between the probe and ship is increasing at a rate of <math> \sqrt{41} \ m/s</math>.

Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i

2012-12-04T03:08:47Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Hint 2

2012-12-04T03:08:07Z

CindyBlois: Created page with "What do the zeros of :<math> \frac{dy}{dt} = y^3-y^2-2y </math> represent?"

What do the zeros of
:<math>
\frac{dy}{dt} = y^3-y^2-2y
</math>
represent?

Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Hint 1

2012-12-04T03:07:28Z

CindyBlois:

Is <math>y </math> increasing or decreasing when <math>t=0 </math>? Using the graph of <math>dy/dt </math> vs. <math>y </math> as the hint in the problem suggests, how can you predict the behaviour of <math>y</math> as time passes?

Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Hint 1

2012-12-04T03:06:49Z

CindyBlois:

Is <math>y </math> increasing or decreasing when <math>t=0 </math>? Using the graph of <math>dy/dt </math> vs. <math>y </math>as the hint in the problem suggests, how can you predict the behaviour of <math>y</math> as time passes?

Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Solution 1

2012-12-04T02:59:45Z

CindyBlois:

Let's follow the hint and sketch a graph of the polynomial on the right hand side. Note that
:<math>\frac{dy}{dt}=y^3-y^2-2y = y( y^2-y-2) = y( y-2)(y+1). </math>
The roots are at 0, 2 and -1.

[[File:M102graph.jpg|thumb|A graph of <math>\frac{dy}{dt} </math> vs. <math>y </math>]]

We are given that y(0)=1, which is in between 0 and 2. In this region, <math>y </math> and <math>y+1 </math> are both positive and <math>y-2 </math> is negative. Therefore, <math> \frac{dy}{dt}<0</math>, which means that <math>y </math> is decreasing (throughout the interval (0,2)). As <math> y</math> approaches zero, we see that <math> \frac{dy}{dt}</math> also approaches zero. (Note that <math>y </math> could never cross zero, since <math> \frac{dy}{dt}=0</math> at the point where <math> y=0</math>. In other words, if <math>y </math> ever hits 0, it stays there!)

Therefore,
:<math>\lim_{t\to\infty} y(t)=0. </math>

File:M102graph.jpg

2012-12-04T02:39:07Z

CindyBlois:

=={{int:filedesc}}==
{{Information
|description={{en|1=A graph of dy/dx vs. y.}}
|date=2012-12-03
|source={{own}}
|author=[[User:CindyBlois|CindyBlois]]
|permission=
|other_versions=
|other_fields=
}}

=={{int:license-header}}==
{{self|cc-by-sa-2.5-ca}}

Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Solution 1

2012-12-04T02:37:22Z

CindyBlois: Created page with " Let's follow the hint and sketch a graph of the polynomial on the right hand side. Note that :<math>\frac{dy}{dt}=y^3-y^2-2y = y( y^2-y-2) = y( y-2)(y+1). </math> The roots..."

Let's follow the hint and sketch a graph of the polynomial on the right hand side. Note that
:<math>\frac{dy}{dt}=y^3-y^2-2y = y( y^2-y-2) = y( y-2)(y+1). </math>
The roots are at 0, 2 and -1.

We are given that y(0)=1, which is in between 0 and 2. In this region, <math>y </math> and <math>y+1 </math> are both positive and <math>y-2 </math> is negative. Therefore, <math> \frac{dy}{dt}<0</math>, which means that <math>y </math> is decreasing.
<math> </math>

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 03

2012-12-04T01:19:09Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH100/December 2010/Question 03/Hint 1

2012-12-04T01:17:54Z

CindyBlois:

Clearly define your variables using diagrams or sentences. Which quantities remain constant and which quantities are changing? What data is given and what are you trying to solve for?

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 03/Hint 1

2012-12-04T01:16:57Z

CindyBlois: Created page with "Clearly define your variables using diagrams or sentences. Which quantities remain constant and which quantities are changing? What do you know and what are you trying to so..."

Clearly define your variables using diagrams or sentences. Which quantities remain constant and which quantities are changing? What do you know and what are you trying to solve for?

Science:Math Exam Resources/Courses/MATH100/December 2010/Question 03/Solution 1

2012-12-04T01:14:56Z

CindyBlois: Created page with "Let <math>V_1 </math>, <math> r_1</math>, and <math> h_1</math> be the volume, radius, and depth of the water in the large pool and let <math>V_2 </math>, <math> r_2</math>, a..."

Let <math>V_1 </math>, <math> r_1</math>, and <math> h_1</math> be the volume, radius, and depth of the water in the large pool and let <math>V_2 </math>, <math> r_2</math>, and <math> h_2</math> be the volume, radius, and depth of the water in the small pool. We know that <math>r_1=8 </math>m and <math> r_2=5</math>m. Since the pools are being filled at the same rate (in <math> m^3/min</math>),

:<math> \frac{dV_1}{dt} = \frac{dV_2}{dt}.</math>

We know that <math> \frac{dh_2}{dt}=0.5 m/min</math> and we want to find <math> \frac{dh_1}{dt}</math>.

By the volume formula for a cylinder,

:<math>V_1 = \pi\, r_1^2 h_1 </math> and <math>V_2 = \pi\, r_2^2 h_2 </math>.

Note that <math> r_1</math> and <math> r_2</math> are constants. The variables of the problem are <math> V_1, V_2, h_1,</math> and <math> h_2. </math>

Since <math> \frac{dV_1}{dt} = \frac{dV_2}{dt}</math>,

:<math>\pi\, r_1^2 \frac{dh_1}{dt} =\pi\, r_2^2 \frac{dh_2}{dt} </math>.

Hence, the water depth in the the larger pool is increasing at a rate of
:<math> \frac{dh_1}{dt} =\frac{r_2^2}{r_1^2 } \frac{dh_2}{dt} = \frac{(5m)^2}{(8m)^2} \frac12 m/min = \frac{25}{128} m/min</math>.

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 03

2012-12-04T00:27:55Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (c)

2012-12-04T00:22:35Z

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Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (n)

2012-11-23T01:20:49Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (n)/Hint 2

2012-11-22T21:34:01Z

CindyBlois:

The linear approximation to a function <math>f</math> at <math>x=a</math> is
:<math>L(x)=f(a)+f'(a)(x-a).</math>

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (n)

2012-11-22T21:32:50Z

CindyBlois:

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Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (n)/Hint 1

2012-11-22T21:31:16Z

CindyBlois:

The graph of the linear approximation taken at <math>a=1</math> is the tangent line to the curve at <math>a=1</math>.

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 05/Solution 1

2012-11-20T08:07:38Z

CindyBlois:

The quantity we want to minimize (the "objective function"), is the profit, <math> P</math>. The profit is equal to the revenue, <math> pq,</math> minus the cost, <math>C(q) </math>.
:<math> P=pq-C(q).</math>
Our constraint is the demand equation, which gives us a relationship between <math> p</math> and <math>q </math>:
:<math>p=400-50q .</math>
We now use the constraint to express the objective function <math> P</math> in terms of one variable only. Since <math> C</math> is already expressed in terms of <math>q </math>, it will be easiest to express <math> P</math> as a function of <math> q</math> rather than <math>p </math>.
:<math>P(q) = (400-50q )q-C(q) = 400q - 50q^2 - C(q) </math>
The domain of interest is <math>[0,8] </math>, because we cannot produce a negative quantity of jackets, and in order to sell more than 8 jackets per month, we'd actually have to start paying people to take them (according to the demand equation). In order to find the absolute maximum value of <math>P(q) </math> in this interval, we start by looking for its critical points.
:<math>P'(q) = 400-100q - C'(q) = 400-100q - \frac{800}{q+5} </math>
This exists everywhere inside the domain. In order to search for its zeros, it is helpful to take out a factor of <math>\frac{100}{q+5} </math> (which is never equal to zero), so that we are left with a polynomial in <math>q </math>. (Hopefully we'll be able to find the roots of that polynomial.)
:<math>\begin{align}P'(q) &= \frac{100}{q+5}\left( (4-q)(q+5) - 8 \right) \\
&= \frac{100}{q+5}\left( 4q +20 -q^2-5q -8 \right)\\
&= -\frac{100}{q+5}\left( q^2+q -12 \right) \\
&= -\frac{100}{q+5} ( q+4)(q-3) \end{align}</math>
This is equal to zero when <math> q</math> is equal to 3 or -4, but only 3 lies in the domain of interest. It is easy to check that the profit is actually maximized when <math>q=3</math> because when <math>q<3,</math> <math> P'(q)>0</math>, and when <math>q>3, </math> <math> P'(q)<0</math>. (Hence, the profit function is increasing when <math>q<3</math> and decreasing when <math>q>3</math>.) Therefore, the profit is maximized when 3 jackets are sold per month.

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 05/Solution 1

2012-11-20T08:06:17Z

CindyBlois:

The quantity we want to minimize (the "objective function"), is the profit, <math> P</math>. The profit is equal to the revenue, <math> pq,</math> minus the cost, <math>C(q) </math>.
:<math> P=pq-C(q).</math>
Our constraint is the demand equation, which gives us a relationship between <math> p</math> and <math>q </math>:
:<math>p=400-50q .</math>
We now use the constraint to express the objective function <math> P</math> in terms of one variable only. Since <math> C</math> is already expressed in terms of <math>q </math>, it will be easiest to express <math> P</math> as a function of <math> q</math> rather than <math>p </math>.
:<math>P(q) = (400-50q )q-C(q) = 400q - 50q^2 - C(q) </math>
The domain of interest is <math>[0,8] </math>, because we cannot produce a negative quantity of jackets, and in order to sell more than 8 jackets per month, we'd actually have to start paying people to take them (according to the demand equation). In order to find the absolute maximum value of <math>P(q) </math> in this interval, we start by looking for its critical points.
:<math>P'(q) = 400-100q - C'(q) = 400-100q - \frac{800}{q+5} </math>
This exists everywhere inside the domain. In order to search for its zeros, it is helpful to take out a factor of <math>\frac{100}{q+5} </math> (which is never equal to zero), so that we are left with a polynomial in <math>q </math>. (Hopefully we'll be able to find the roots of that polynomial.)
:<math>P'(q) = \frac{100}{q+5}\left( (4-q)(q+5) - 8 \right) </math>
:<math> = \frac{100}{q+5}\left( 4q +20 -q^2-5q -8 \right)</math>
:<math> = -\frac{100}{q+5}\left( q^2+q -12 \right) </math>
:<math> = -\frac{100}{q+5} ( q+4)(q-3) </math>
This is equal to zero when <math> q</math> is equal to 3 or -4, but only 3 lies in the domain of interest. It is easy to check that the profit is actually maximized when <math>q=3</math> because when <math>q<3,</math> <math> P'(q)>0</math>, and when <math>q>3, </math> <math> P'(q)<0</math>. (Hence, the profit function is increasing when <math>q<3</math> and decreasing when <math>q>3</math>.) Therefore, the profit is maximized when 3 jackets are sold per month.

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 04/Solution 1

2012-11-20T08:00:08Z

CindyBlois: Just removed comment about setting A'(q)=0 because it is mentioned later and because we also call points where the derivative doesn't exist critical points in this course. Added note about q>0 for taking fourth root.

Let <math>A(q)</math> denote the average cost per kilogram of producing <math>q</math> kilograms of this chemical,
:<math>A(q) = \frac{C(q)}{q} = 3q^{1/3}+50+\frac{10000}{q}.</math>
The domain of interest is <math>[0,\infty)</math> because the company cannot produce negative kilograms of their product. In order to find the absolute minimum of the average cost function, we begin by looking for its critical points. The derivative of <math>A</math> is
:<math>A'(q) = q^{-2/3}-\frac{10000}{q^2} = \frac{1}{q^2} (q^{4/3}-10000)</math>
This exists everywhere in the interior of our domain and is equal to zero when <math> q^{4/3}-10000 =0</math>, that is, when
:<math>q^{\frac43}=10000 = 10^4.</math>
Since <math>q\geq 0</math>, we can take the positive fourth root of both sides to obtain
:<math>q^{\frac13}=10.</math>
Hence,
:<math>q=10^3 = 1000.</math>

It is easy to check that this is indeed an absolute minimum by noting that <math> A'(q)<0</math> when <math>q<1000 </math> and <math> A'(q)>0</math> when <math>q>1000 </math>. Since <math>A(q) </math> is decreasing when <math> q<1000</math> and increasing when <math> q>1000</math>, <math>A(q) </math> must have an absolute minimum at <math> q=1000.</math>

Hence, the company should produce 1000 kg of chemical in order to minimize its average cost.

Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (n)/Solution 1

2012-11-20T07:40:48Z

CindyBlois:

We use the linear approximation of <math>\ln(x)</math> at <math>x=1</math> to estimate <math>\ln(0.8)</math>. (Approximating at ''x = 1'' is a logical choice, as 1 is close to 0.8, and we know that <math>\ln(1) = 0</math>.)

We know <math>\frac{d}{dx}\ln(x)=\frac1x</math>, which is equal to 1 when <math>x=1</math>. Substituting this into the linear approximation formula, we get,
:<math>\ln(0.8)\approx \ln(1)+1(0.8-1) = 0-0.2=0.2</math>