Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 12 (b)

MATH 180 December 2017

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Question 12 (b)
Suppose you wish to construct an enclosed box of maximal volume out of an ${\textstyle L\times L}$ square sheet of cardboard by removing the shaded areas below and folding along the dotted lines. (Each unshaded region will form exactly one side of the box.) (b) Determine the three dimensions (length, width and height) of the box of maximal volume.

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
Find $x$ which maximizes the volume $V(x)$ given in part (a).

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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. First, we find $0\leq x\leq L$ which maximizes the volume given in part (a) by $$V(x)=(L-2x)\left({\frac {L}{2}}-x\right)x={\frac {1}{2}}(L-2x)^{2}x$$ . To find its critical point, we take a derivative. Using the product and chain rules, we get $${\begin{aligned}V'(x)&={\frac {1}{2}}((L-2x)^{2}x)'={\frac {1}{2}}\left(2(L-2x)(-2)x+(L-2x)^{2}\right)\\&={\frac {1}{2}}(L-2x)(-4x+L-2x)={\frac {1}{2}}(L-2x)(L-6x).\end{aligned}}$$ . Therefore, we have $V'(x)=0$ if $L-2x=0$ and $L-6x=0$. In other words, the critical points are $x={\frac {L}{2}}$ and $x={\frac {L}{6}}$. Plugging these values into $V(x)$, we obtain $V\left({\frac {L}{2}}\right)=0$ and $$V\left({\frac {L}{6}}\right)={\frac {1}{2}}\left(L-2\cdot {\frac {L}{6}}\right)^{2}{\frac {L}{6}}={\frac {1}{27}}L^{3}.$$ Obviously, the maximum is obtained when $x={\frac {L}{6}}$. To find the corresponding length and width, we recall their relations with $x$ in part (a): $$l=L-2\cdot {\frac {L}{6}}={\frac {2}{3}}L,$$ $$w={\frac {L}{2}}-{\frac {L}{6}}={\frac {1}{3}}L.$$ As a result, we have the maximum volume of the box when the length, width, and height are given by $\color {blue}{\frac {2}{3}}L,\ {\frac {1}{3}}L,\ {\frac {1}{6}}L$, respectively.