Question

If \(1200\,c{m^2}\)of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Short answer

The largest possible volume of the box is \(4000\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\).

Step-by-step solution

Second Derivative test

If \(f'\left( c \right) = 0\), then the value of \(f\left( x \right)\) is maximum at \(x = c\) if \(f''\left( c \right) < 0\) and the value of \(f\left( x \right)\) is minimum at \(x = c\) if \(f''\left( c \right) > 0\).

Optimization of the volume of the box with a square base

Let the base of the box be \(b\) and the height of the box be \(h\). Now, the box has an open-top. So, the surface area of the box will be \(S = {b^2} + 4bh\).

Given that the surface area is \(1200\,{\rm{c}}{{\rm{m}}^{\rm{2}}}\). So, \(1200 = {b^2} + 4bh \Rightarrow h = \frac{{1200 - {b^2}}}{{4b}}\).

Now, the volume of the box will be equal to \(V = {b^2}h\). Put \(h = \frac{{1200 - {b^2}}}{{4b}}\) in the above equation:

\(\begin{aligned}{c}V = {b^2}\left( {\frac{{1200 - {b^2}}}{{4b}}} \right)\\ = 300b - \frac{{{b^3}}}{4}\end{aligned}\)

Differentiate the volume with respect to \(b\) optimizing it:

\(\begin{aligned}{l}V' = \frac{d}{{db}}\left( {300b - \frac{{{b^3}}}{4}} \right)\\V' = 300 - \frac{3}{4}{b^2}\end{aligned}\)

Find critical points as follows:

\(\begin{aligned}{c}V' = 0\\300 - \frac{3}{4}{b^2} = 0\\300 = \frac{3}{4}{b^2}\\{b^2} = 400\\b = 20\end{aligned}\)

The second derivative of the volume function \(V'' = - \frac{3}{2}b\) is always less than zero. Therefore, the volume is maximum at \(b = 20\).

At \(b = 20\), the value of \(h\) is equal to:

\(\begin{aligned}{c}h = \frac{{1200 - {{20}^2}}}{{4\left( {20} \right)}}\\ = 10\end{aligned}\)

Hence, the largest possible volume of the box is \({b^2}h = {\left( {20} \right)^2} \cdot 10 = 4000\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\).