What is the radius of a cylindrical soda can with volume of 512 cubic inches that will use the minimum material?
$\displaystyle 512 = \pi r^2 h$
$\displaystyle h = \frac{512}{\pi r^2}$
material required = total surface area of the cylinder ...
$\displaystyle A = 2\pi r^2 + 2\pi rh$
sub in the expression for $\displaystyle h$ given above to get $\displaystyle A$ in terms of $\displaystyle r$ only ...
find $\displaystyle \frac{dA}{dr}$ and minimize