Science:Math Exam Resources/Courses/MATH110/April 2011/Question 07 (c)

MATH110 April 2011

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Question 07 (c)
As long as $m<0$, the line you gave as an answer in part (a) cuts out a triangle in the upper righthand quadrant, whose vertices are at the origin, the x-intercept, and the y-intercept. Find m so that the area of this triangle is as small as possible.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Draw a picture of a line and label the sides of the triangle that is formed. How could you write each side in terms of $m$?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

The question is asking us to minimize the area of a triangle. So the function that we will be minimizing is $A={\frac {1}{2}}bh$ where b is the triangle's base and h is its height. In order to optimize (in this case, minimize), we need to reduce this formula to a function of one variable. If we draw a picture of a line with negative slope that passes through (1,1), we will see the triangle formed by the axes and the line and can label its base and height. So the base of the triangle is equal to the distance between 0 and the x-intercept of the line (which is simply the value of the x-intercept) and similarly, the height is equal to the y-intercept. We calculated the x and y intercept in the previous part of the question, so now we can simply plug them into our area formula for b and h. ${\begin{aligned}A&={\frac {1}{2}}\left({\frac {m-1}{m}}\right)(-m+1)\\&={\frac {-m^{2}+2m-1}{2m}}\end{aligned}}$ We now take the derivative of A... ${\begin{aligned}A'&={\frac {(-2m+2)(2m)-(-m^{2}+2m-1)(2)}{4m^{2}}}\\&={\frac {-2(m-1)(m)+(m-1)^{2}}{2m^{2}}}\\&={\frac {(m-1)(-2m+m-1)}{2m^{2}}}\\&={\frac {(m-1)(-m-1)}{2m^{2}}}\end{aligned}}$ ...and find its critical points. The derivative is undefined when $m=0$. Setting it equal to zero: $0={\frac {(m-1)(-m-1)}{2m^{2}}}$ $0=(m-1)(-m-1)$ yields additional critical points of $m=-1,$ and $m=1$. However, as stated in the question, $m<0$, so the only valid critical point is $m=-1$. To check that this is a minimum, you could use the first or second derivative test.

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