Question

A box with a square base and open top must have a volume of\(32000\,c{m^3}\). Find the dimensions of the box that minimize the amount of material used.

Short answer

The dimensions of the box that minimize the amount of material used are \(40 \times 40 \times 20\).

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 material used to make a box with a square base

Let the base of the box be \(b\) and the height of the box be \(h\). Now, the volume of the box is \(V = {b^2}h\). Given that the volume of the box is \(32000\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\). So, \(\begin{aligned}{l}32000 = {b^2}h\\ \Rightarrow h = \frac{{32000}}{{{b^2}}}\end{aligned}\).

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

Put \(h = \frac{{32000}}{{{b^2}}}\) in the above equation:

\(\begin{aligned}{c}S = {b^2} + 4b\left( {\frac{{32000}}{{{b^2}}}} \right)\\ = {b^2} + \frac{{128000}}{b}\end{aligned}\)

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

\(\begin{aligned}{l}S' = \frac{d}{{db}}\left( {{b^2} + \frac{{128000}}{b}} \right)\\S' = 2b - \frac{{128000}}{{{b^2}}}\end{aligned}\)

Find critical points as follows:

\(\begin{aligned}{c}S' = 0\\2b - \frac{{128000}}{{{b^2}}} = 0\\2b = \frac{{128000}}{{{b^2}}}\\{b^3} = 64000\\b = 40\end{aligned}\)

The second derivative of the surface area function \(S'' = 2 + \frac{{256000}}{{{b^3}}}\) is always greater than zero. Therefore, the surface area is minimum at \(b = 40\).

At \(b = 40\), the value of \(h\) is equal to: \(h = \frac{{32000}}{{{{\left( {40} \right)}^3}}} = 20\).

Hence, the dimensions of the box that minimize the amount of material used are \(40 \times 40 \times 20\).