Question

Rework Exercise 21 assuming the container has a lid that is made from the same material as the sides.

Short answer

The cost of material for the least expensive such container is \(\$ 191.28\).

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 side’s costs $6 per square meter. Now, the cost of the material to make the box with a lid will be: \(C = 10\left( {2{x^2}} \right) + 6\left[ {2\left( {2xy} \right) + 2\left( {xy} \right) + 2{x^2}} \right] = 32{x^2} + 36xy\).

Put \(y = \frac{5}{{{x^2}}}\) in the above equation to get \(C = 32{x^2} + 36x\left( {\frac{5}{{{x^2}}}} \right) = 32{x^2} + \frac{{180}}{x}\).

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

\(\begin{aligned}{l}C' = \frac{d}{{dx}}\left( {32{x^2} + \frac{{180}}{x}} \right)\\C' = 64x - \frac{{180}}{{{x^2}}}\end{aligned}\)

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

\(\begin{aligned}{c}C' = 0\\64x - \frac{{180}}{{{x^2}}} = 0\\64x = \frac{{180}}{{{x^2}}}\\{x^3} = \frac{{180}}{{64}}\\x = \sqrt[3]{{\frac{{45}}{{16}}}}\end{aligned}\)

Now, it is clear that the second derivative \(C'' = 64 + \frac{{360}}{{{x^3}}}\) is always greater than zero. So, \(x = \sqrt[3]{{\frac{{45}}{{16}}}}\) is the absolute minimum for \(C\).

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

\(\begin{aligned}{c}C\left( {\sqrt[3]{{\frac{{45}}{{16}}}}} \right) = 64{\left( {\sqrt[3]{{\frac{{45}}{{16}}}}} \right)^2} + \frac{{180}}{{\sqrt[3]{{\frac{{45}}{{16}}}}}}\\ \approx 191.28\end{aligned}\)

Hence, the cost of material for the least expensive such container is \(\$ 191.28\).