Let r and h be the radius and height of the cylinder. Then, A = 2πrh + $2\pi r^2$ (Given) ⇒ h = $\frac{A-2\pi r^2}{2\pi r}$ Now, Volume (V) = $\pi r^2 h$ ⇒ V = $\pi r^2 \left(\frac{A-2\pi r^2}{2\pi r}\right)$ = $\frac{1}{2}\left(Ar - 2\pi r^3\right)$ ⇒ $\frac{dV}{dr}$ = $\frac{1}{2}\left(A - 6\pi r^2\right)$ ... (1) ⇒ $\frac{d^2V}{dr^2}$ = $\frac{1}{2} - 12\pi r$ ... (2) Now, $\frac{dV}{dr}$ = 0 ⇒ $\frac{1}{2}\left(A - 6\pi r^2\right)$ = 0 ⇒ $r^2$ = $\frac{A}{6\pi}$ ⇒ r = $\sqrt{\frac{A}{6\pi}}$ Now, $\left|\frac{dV^2}{dr^2}\right|_{r=\sqrt{\frac{A}{6\pi}}}$ = $\frac{1}{2}\left(-12\pi\sqrt{\frac{A}{6\pi}}\right)$ < 0 Therefore, Volume is maximum at $r = \sqrt{\frac{A}{6\pi}}$ ⇒ $r^2$ = $\frac{A}{6\pi}$ ⇒ $6\pi r^2$ = A ⇒ $2\pi r^2$ = 2πrh + $4\pi r^2$ ⇒ $4πr^2$ = 2πrh ⇒ 2r = h Hence, the volume is maximum if its height is equal to its diameter.