Let r and h be the radius and height of the right circular cylinder with the open top.
So surface area of the cylinder S is given by,
S = $\mathit{\pi }{\mathit{r}}^2+2\mathit{\pi }\mathit{r}\mathit{h}$
⇒ h = $\frac{\mathit{S}-\mathit{\pi }{\mathit{r}}^2}{2\mathit{\pi }\mathit{r}}$ ... (i)
Let V be the volume, so
V = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$ = $\mathit{\pi }{\mathit{r}}^2\frac{\left(\mathit{S}-\mathit{\pi }{\mathit{r}}^2\right)}{2\mathit{\pi }\mathit{r}}$ = $\mathit{r}\frac{\left(\mathit{S}-\mathit{\pi }{\mathit{r}}^2\right)}{2}$
$\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = $\frac{\mathit{S}}{2}-\frac{3\mathit{\pi }{\mathit{r}}^2}{2}$ ... (ii)
for maxima or minima $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = 0
⇒ S = $3\mathit{\pi }{\mathit{r}}^2$ or r = $\sqrt{\frac{\mathit{S}}{3\mathit{\pi }}}$
Using this (i)
h = $\frac{2\mathit{\pi }{\mathit{r}}^2}{2\mathit{\pi }\mathit{r}}$ = r
$\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{r}}^2}$ = - 3 πr
= - 3π $\sqrt{\frac{\mathit{S}}{3\mathit{\pi }}}$ < 0
So, r = $\sqrt{\frac{\mathit{S}}{3\mathit{\pi }}}$ is a point of maxima
And in this case radius of base = height