A box with square base and no top is to hold a volume \(100\). Find the dimensions of the box that requires the least material for the five sides.

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The dimensions that will result in "least material" for the five sides are dimensions that result in "smallest surface area". So, the quantity that we have to "minimize" is the surface area, \(S\).
We have to "minimize" the surface area, \(S\).

\(S=x^2 +4xy\)

Remember, no top!

Next, we express \(S\) as a function of, say, \(x\). In order to do that, we have to express \(y\) in terms of \(x\).

We know that \(V=100\). Therefore,

\(x^2\cdot \answer { y}=100\), and

\(y=\frac {100}{\answer {x^2}}\).

It follows that

\(S(x)=x^2 +\frac {400}{\answer {x}}\).

So, we have to find the (global) minimum of \(S\) on its domain, \((0,\answer {\infty })\). We have to find critical points of \(S\). Therefore, we have to compute \(S'(x)\).
\(S'(x)=2x-\frac {400}{\answer {x^2}}\).
We have to find critical points of \(S\). So, we have to solve the equation

\(2x-\frac {400}{\answer {x^2}}=0\).

It follows that \(S\) has the only critical point \(x=\answer {200^{\frac {1}{3}}}\).

Since

\(S'(x)=2x-\frac {400}{x^{2}}=\frac {2(x^{3}-200)}{x^{2}}\), it follows that

\(S'(x)<0\) on \((0, 200^{\frac {1}{3}})\) and that \(S'(x)>0\) on \((200^{\frac {1}{3}},\infty )\).

This means that \(S\) is decreasing on \((0, 200^{\frac {1}{3}})\) and increasing on \((200^{\frac {1}{3}},\infty )\). That means that the function \(S\) has both a local minimum and global minimum at \(x=c\).

So, the dimensions that minimize the surface area are

\[ \text {width}=\answer {(2) 5^{2/3}},\qquad \text {length}=\answer {(2) 5^{2/3}},\qquad \text {height}=\answer {5^{2/3}} \]