Answer
$V = \frac{{64000}}{\pi }$ $c{m^3}$ is the maximum volume of the cylindrical soda can.
The dimensions of the can when the volume is maximum:
$\left( {r,h} \right) = \left( {\frac{{40}}{\pi },40} \right)$ cm,
where $r$ and $h$ are the radius and the height of the can, respectively.
Work Step by Step
Let $r$ and $h$ be the radius and the height of the can. The sum of its height and its circumference is fixed $s = h + 2\pi r = 120$ cm.
The volume $V$ of the cylindrical soda can is $\pi {r^2}h$.
Using $s = h + 2\pi r = 120$, write the volume $V$:
$V\left( r \right) = \pi {r^2}\left( {120 - 2\pi r} \right) = 120\pi {r^2} - 2{\pi ^2}{r^3}$
To find the maximum of $V$ we solve the equation $V' = 0$ and find the critical points:
$V' = 240\pi r - 6{\pi ^2}{r^2} = 0$
$r\left( {240\pi - 6{\pi ^2}r} \right) = 0$
The solutions are $r=0$ and $r = \frac{{40}}{\pi }$. Since $r \ne 0$, so the critical point is at $r = \frac{{40}}{\pi }$ cm.
Using $h + 2\pi r = 120$, we obtain $h=40$ cm.
We can verify that these dimensions $\left( {r,h} \right) = \left( {\frac{{40}}{\pi },40} \right)$ maximize the volume of the can by examining the second derivatives:
$V{\rm{''}} = 240\pi - 12{\pi ^2}r$
$V{\rm{''}}\left( {\frac{{40}}{\pi }} \right) = 240\pi - 480\pi = - 240\pi $
Since $V{\rm{''}}\left( {\frac{{40}}{\pi }} \right) < 0$, so $V\left( {\frac{{40}}{\pi }} \right) = \frac{{64000}}{\pi }$ $c{m^3}$ is the maximum volume of the cylindrical soda can.
