Soup Can Solution

A large soup can is to be designed so that the can will hold 
36 cubic inches of soup.

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Find the values of $x$ and $h$ for which the amount of metal
needed is as small as possible.

What is the value of $x$?

What is the value of $h$?
The soup can consists of metal for a top, a bottom, and the piece that wraps around for the side. The total amount of metal used is
$$C=\pi x^2+\pi x^2+2\pi xh=2\pi x^2+2\pi xh$$
where both $h$ and $x$ must be positive numbers. Since the total volume of the soup can must be 36 cubic inches, it follows that
$$V=\pi x^2 h=36$$
from which one may solve for $h$ to get
$$h={36\over\pi x^2}$$ .
Substitute into the expression for the total amount of metal.
$$C=2\pi x^2+{2\pi x}\Big({36\over\pi x^2}\Big) =2\pi x^2+72 x^{-1}$$
Differentiate.
$$C'=4\pi x-72 x^{-2}=4\pi x-{72\over x^2}$$
Set the derivative equal to zero and solve for $x$ .
$$4\pi x-{72\over x^2}=0$$
$$x^3={72\over 4\pi}$$
Hence
$$x=\Big({72\over 4\pi}\Big)^{1/3}=1.789400458$$ inches
and
$$h={36\over\pi (1.789400458)^2}=3.578800913$$ inches
This is indeed the dimensions for minimum amount of metal since $C$ becomes infinite as $x\to 0$ and also as $x\to\infty$ .