Let r and h be the radius and height of the cylinder. Then,
A = 2πrh + $2\mathit{\pi }{\mathit{r}}^2$ (Given)
⇒ h = $\frac{\mathit{A}-2\mathit{\pi }{\mathit{r}}^2}{2\mathit{\pi }\mathit{r}}$
Now, Volume (V) = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$
⇒ V = $\mathit{\pi }{\mathit{r}}^2\left(\frac{\mathit{A}-2\mathit{\pi }{\mathit{r}}^2}{2\mathit{\pi }\mathit{r}}\right)$ = $\frac{1}{2}\left(\mathit{A}\mathit{r}-2\mathit{\pi }{\mathit{r}}^3\right)$
⇒ $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = $\frac{1}{2}\left(\mathit{A}-6\mathit{\pi }{\mathit{r}}^2\right)$ ... (1)
⇒ $\frac{{\mathit{d}}^2\mathit{V}}{\mathit{d}{\mathit{r}}^2}$ = $\frac{1}{2}-12\mathit{\pi }\mathit{r}$ ... (2)

Now, $\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$ = 0 ⇒ $\frac{1}{2}\left(\mathit{A}-6\mathit{\pi }{\mathit{r}}^2\right)$ = 0
⇒ ${\mathit{r}}^2$ = $\frac{\mathit{A}}{6\mathit{\pi }}$ ⇒ r = $\sqrt{\frac{\mathit{A}}{6\mathit{\pi }}}$
Now, ${\left|\frac{\mathit{d}{\mathit{V}}^2}{\mathit{d}{\mathit{r}}^2}\right|}_{\mathit{r}=\sqrt{\frac{\mathit{A}}{6\mathit{\pi }}}}$ = $\frac{1}{2}\left(-12\mathit{\pi }\sqrt{\frac{\mathit{A}}{6\mathit{\pi }}}\right)$ < 0
Therefore, Volume is maximum at $\mathit{r}=\sqrt{\frac{\mathit{A}}{6\mathit{\pi }}}$
⇒ ${\mathit{r}}^2$ = $\frac{\mathit{A}}{6\mathit{\pi }}$ ⇒ $6\mathit{\pi }{\mathit{r}}^2$ = A
⇒ $6\mathit{\pi }{\mathit{r}}^2$ = 2πrh + $2\mathit{\pi }{\mathit{r}}^2$
⇒ $4\mathit{\pi }{\mathit{r}}^2$ = 2πrh ⇒ 2r = h
Hence, the volume is maximum if its height is equal to its diameter.