Minimize the cost of a box of fixed volume if the sides are twice as expensive as the base and top

Question

Suppose, to build a box (a rectangular solid) of fixed volume and square base, the cost per square inch of the base and top is twice that of the four sides. Choose dimensions for the box that minimize the cost of building it, if the volume is (say) $2000$ cubic inches.

Here is what I have:

Let $x$ be the length (in.) of a side of the base. Let $h$ be the height (in.). Volume is given by $$V = x^2h = 2000\;.$$ Furthermore let $m$ be the cost (per square in.) of the base and top, and $n$ the cost (per square in.) of the sides. Then total cost, $C$, is given by $$C = m(2x^2) + n(4xh)\;.$$

We also know $$m = 2n\;.$$

To do: Express $C$ in terms of one variable, then use standard elementary calculus techniques to minimize $C$. I am stuck at this point: How do I get $C$ in terms of a single variable? I can substitute $h = 2000/x^2$, and either $m = 2n$ or $n = m/2$, but this leaves $C$ in terms of $x$ and $n$ or $m$.

Answer 1

You are off to a good start. If $C(x)$ represents the cost of building a box with a square base of side length $x$, then (assuming that your computations are correct–I did not check them) $$ C(x) = m(2x^2) + n(4xh) = 4nx^2 + \frac{8000 n}{x} = 4n \left(x^2 + \frac{2000}{x} \right). $$ This tends to infinity both as $x\to 0^+$ and as $x\to +\infty$, but is otherwise continuously differentiable, hence (by Fermat's theorem) we expect the minimum cost to correspond to the value of the cost function $C$ at a critical number. The critical numbers can be obtained by determining when the derivative of $C$ is zero, i.e. we need to solve: \begin{align} &0 = C'(x) = \frac{\mathrm{d}}{\mathrm{d}x} 4n \left(x^2 + \frac{2000}{x} \right) = 4n \left( 2x - \frac{2000}{x^2} \right) \\ &\qquad\implies 0 = 2x - \frac{2000}{x^2} \\ &\qquad\implies 0 = 2x^3 - 2000 \\ &\qquad\implies x^3 = 1000 \\ &\qquad\implies x = \sqrt[3]{1000} = 10. \end{align} As there is only one critical number, it must be that this corresponds to the minimum (though it wouldn't be a bad idea to confirm this with, for example, the first derivative test). Note that the actual cost plays no role in determining where the minimum occurs. It will change the value of that minimum, but not the location. I expect that you should be able to finish the problem from here.

Answer 2

You got $C(x,h) = 4n(x^2 + xh)$ and we know that $V = x^2h$ is fixed. Eliminating $h$ gives:

$$C(x) = 4n\left(x^2 + \frac{V}x\right)$$

Now we can use the AM-GM inequality:

$$C(x) = 4n\left(x^2 + \frac{V}x\right) = 4n\left(x^2 + \frac{V}{2x} + \frac{V}{2x}\right) \ge 4n \cdot 3\sqrt[3]{x^2 \cdot \frac{V}{2x} \cdot \frac{V}{2x}} = 12n \sqrt[3]{\frac{V^2}4}$$

Equality is attained for $x^2 = \frac{V}{2x}$, or $x^3 = \frac{V}2 = 1000$ which gives $x = 10$ and then $h = 20$.