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

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MATH 180 December 2017

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• December 2017 •

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.) $A box$ A box (b) Determine the three dimensions (length, width and height) of the box of maximal volume.

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

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Solution
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.