Let $\mathrm{S}$ be the given surface area of the closed cylinder whose radius is $r$ and height $h$ let $v$ be the its Volume. Then Surface area $\mathrm{S}=2 \pi \mathrm{r}^2+2 \pi \mathrm{rh}, \mathrm{h}=\frac{\mathrm{S}-2 \pi \mathrm{r}^2}{2 \pi \mathrm{r}} \quad \ldots(i)$


$\therefore$ Volume $\mathrm{V}=\pi \mathrm{r}^2 \mathrm{~h}$
$=\pi r^2\left(\frac{S-2 \pi r^2}{2 \pi r}\right)$
$=\frac{1}{2}\left[\mathrm{Sr}-2 \pi \mathrm{r}^3\right]$
$\therefore \frac{\mathrm{dV}}{\mathrm{dx}}=\frac{1}{2} \times\left[\mathrm{S}-6 \pi \mathrm{r}^2\right] \quad \ldots(ii)$
For maxima and minima $\frac{\mathrm{dV}}{\mathrm{dx}}=0$
$\therefore S=6 \pi r^2 \Rightarrow r=\sqrt{\frac{S}{6 \pi}}$
from (i), $\mathrm{h}=\frac{\mathrm{S}-2 \pi \mathrm{r}^2}{2 \pi \mathrm{r}}$
(Putting values of S)
Now, $\frac{\mathrm{d}^2 \mathrm{v}}{\mathrm{dr}^2}=-6 \pi \mathrm{r} < \mathrm{o}$ at $\mathrm{r}=\sqrt{\frac{\mathrm{s}}{6 \pi}}$
$\therefore \mathrm{V}$ is maximum. Thus, volume is maximum when $\mathrm{h}=2 \mathrm{r}$ i.e. when height of cylinder $=$ diameter of the base.