Question

A rectangular storage container without a lid is to have a volume of \(10\,{{\rm{m}}^{\rm{3}}}\). The length of its base is twice the width. Material for the base costs \(10 per square meter. Material for the sides costs \)6 per square meter. Find the cost of materials for the least expensive such container.

Short answer

The cost of material for the least expensive such container is \(\$ 163.54\).

Step-by-step solution

Assumptions of variables

Let the width of the base is \(x\,{\rm{cm}}\) and height be \(y\,{\rm{cm}}\). So, the length will be \(2x\,{\rm{cm}}\). The image given below describes the given scenario.

Optimization of the cost of the material

For the given box, the volume will be equal to \(V = lwh = \left( {2x} \right)\left( x \right)\left( y \right) = 2{x^2}y\). It is given that volume is \(10\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\). Then, \(10 = 2{x^2}y \Rightarrow y = \frac{5}{{{x^2}}}\).

Since the cost of material for the base is $10 per square meter and material for the sides costs $6 per square meter. Now, the cost of the material to make the box without a lid will be: \(C = 10\left( {2{x^2}} \right) + 6\left( {2\left( {2xy} \right) + 2\left( {xy} \right)} \right) = 20{x^2} + 36xy\).

Put \(y = \frac{5}{{{x^2}}}\) in the above equation to get \(C = 20{x^2} + 36x\left( {\frac{5}{{{x^2}}}} \right) = 20{x^2} + \frac{{180}}{x}\).

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

\(\begin{aligned}{l}C' = \frac{d}{{dx}}\left( {20{x^2} + \frac{{180}}{x}} \right)\\C' = 40x - \frac{{180}}{{{x^2}}}\end{aligned}\)

Find critical points by taking \(C' = 0\):

\(\begin{aligned}{c}C' = 0\\40x - \frac{{180}}{{{x^2}}} = 0\\40x = \frac{{180}}{{{x^2}}}\\{x^3} = \frac{{180}}{{40}}\\x = \sqrt(3){{\frac{9}{2}}}\end{aligned}\)

Now, it is clear that the second derivative \(C'' = 40 + \frac{{360}}{{{x^3}}}\) is always greater than zero. So, \(x = \sqrt(3){{\frac{9}{2}}}\) is the absolute minimum for \(C\).

So, the minimum cost of the material is as follows:

\(\begin{aligned}{c}C\left( {\sqrt(3){{\frac{9}{2}}}} \right) = 20{\left( {\sqrt(3){{\frac{9}{2}}}} \right)^2} + \frac{{180}}{{\sqrt(3){{\frac{9}{2}}}}}\\ \approx 163.54\end{aligned}\)

Hence, the cost of material for the least expensive such container is \(\$ 163.54\).