Question

A package to be mailed using the US postal service may not measure more than 108 inches in length plus girth. (Length is the longest dimension and girth is the largest distance around the package, perpendicular to the length.) Find the dimensions of the rectangular box with square base of greatest volume that may be mailed.

Short answer

The maximum volume of the box is \(11664\,{\rm{i}}{{\rm{n}}^{\rm{3}}}\).

Step-by-step solution

Assumptions of variables

It is given that the length plus girth of the box must not measure more than 108 in. Let the side of the square base is \(x\,{\rm{in}}\) and the length of the box be \(y\,{\rm{in}}\). Then, \(4x + y = 108 \Rightarrow y = 108 - 4x\).

Optimization of the volume

The volume of the box will be equal to \(V = {x^2}y\). Substitute \(y = 108 - 4x\) in the above equation to get \(V = {x^2}\left( {108 - 4x} \right) = 108{x^2} - 4{x^3}\).

Differentiate \(V\) with respect to \(x\) as follows:

\(V' = 216x - 12{x^2}\)

Put \(V' = 0\) to find critical points:

\(\begin{aligned}{c}V' = 0\\216x - 12{x^2} = 0\\18x - {x^2} = 0\\x\left( {18 - x} \right) = 0\end{aligned}\)

The above equation implies that \(x = 0\) and \(x = 18\). Neglect \(x = 0\).

Since the second derivative \(V'' = 216 - 24x\) is less than zero \(x = 18\). Therefore, by the second derivative test, we can say that \(x = 18\) is the point of maximum.

So, the volume of the box can be obtained as follows:

\(\begin{aligned}{c}V = 108{\left( {18} \right)^2} - 4{\left( {18} \right)^3}\\ = 34992 - 23328\\ = 11664\end{aligned}\)

Hence, the maximum volume of the box is \(11664\,{\rm{i}}{{\rm{n}}^{\rm{3}}}\).