Science:Math Exam Resources/Courses/MATH102/December 2019/Question 12

MATH102 December 2019

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Question 12
A 90cm piece of wire is used to create an equilateral triangle, and a rectangle that is twice as long as it is wide. Determine whether and where to cut the wire such that the sum of the areas of the two shapes is maximized. Note: the area of an equilateral triangle with side length $\ell$ is ${\frac {\sqrt {3}}{4}}\ell ^{2}$. As always, completely justify your answer.

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
If $x$ cm of wire is used to create the triangle, $90-x$ is used to make the rectangle. Then set up the equation for the perimeter to get expressions for the side lengths of the triangle and rectangle. Remember to check the area at $x=0$ and $x=90$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Denote the position where we cut the wire as $x$. We construct the triangle with length $x$ and the rectangle with length $90-x$. Note that the perimeter of the triangle is $x$. Therefore, the length of its side is $x/3$. The area of the triangle is ${\frac {\sqrt {3}}{4}}\left({\frac {x}{3}}\right)^{2}$. Similarly, let $\ell$ be length of the shorter side of the rectangle. Then, the perimeter is $2\ell +4\ell =6\ell =90-x$. The length is $\ell ={\frac {90-x}{6}}$ and its area is $2\left({\frac {90-x}{6}}\right)^{2}$. The area to be maximized can be written as $A(x)={\frac {\sqrt {3}}{4}}\left({\frac {x}{3}}\right)^{2}+2\left({\frac {90-x}{6}}\right)^{2}$. We compare its area at the critical point and the end points. Differentiating the area, ${\frac {dA}{dx}}={\frac {2{\sqrt {3}}}{36}}x-4\left({\frac {90-x}{36}}\right)$. Solving for the critical point gives us $2{\sqrt {3}}x-360+4x\implies x=360/(2{\sqrt {3}}+4)$. However, note that this function attains its local minimum at the critical point. Checking values at the end points $A(0){\text{ and }}A(90)$, we can see that the total area is maximized when $x=0$, so when all the wire is used to create the rectangle.