Question

Find the maximum volume of a cylindrical soda can such that the sum of its height and its circumference is 120 cm.

Answer 1

The volume of such a can with base radius $r$ and height $h$ would be

$V(r,h)=\pi r^2h$


We desire the base's circumference and the can's height to add to 120, i.e.

$2\pi r+h=120\implies h=120-2\pi r$

Substituting this into $V(r,h)$ allows us to reduce the volume to a function of a single variable $r$:

$V(r,120-2\pi r)=V(r)=\pi r^2(120-2\pi r)=120\pi r^2-2\pi^2r^3$

Taking the derivative with respect to $r$ yields

$V'(r)=240\pi r-6\pi^2r^2$

Set this equal to 0 and find any critical points:

$240\pi r-6\pi^2r^2=0\implies 6\pi r(40-\pi r)=0\implies r=0\text{ or }r=\dfrac{40}\pi$

This suggests the can will have maximum volume when its radius is $\dfrac{40}\pi\approx12.7$ cm, which would give a volume of about 20,371 sq. cm.