Question

Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

Answer 1

Let $r$ and $h$ be radius and height of given cylinder of surface area $S$.
If $V$ be the volume of cylinder then
$V=\pi r^{2}h$
$V=\frac{\pi r^2.(S-2\pi r^2)}{2\pi r}$ $[\therefore S=2\pi r^2+2\pi rh\Rightarrow \frac{S-2\pi r^2}{2\pi r}=h]$

$\Rightarrow V=\frac{Sr-2\pi r^3}{2}$

$\frac{dV}{dr}=\frac{1}{2}(S-6\pi r^2)$

For maximum or minimum value of $V$
$\frac{dV}{dr}=0$
$\Rightarrow \frac{1}{2}(S-6\pi r^2)=0\Rightarrow S-6\pi r^2=0$
$\Rightarrow r^2=\frac{S}{6\pi }\Rightarrow r=\sqrt{\frac{S}{6\pi }}$

Now, $\frac{d^{2}V}{d{r}^2}=-\frac{1}{2}\times 12\pi r$
$\Rightarrow \frac{d^{2}V}{d{r}^2}=6\pi r$
$\Rightarrow {\left[\frac{d^{2}V}{d{r}^2}\right]}_{r=\sqrt{\frac{S}{6\pi }}}=-ve$

Hence for $r=\sqrt{\frac{S}{6\pi }}$, volume $V$ is maximum.
$\Rightarrow h=\frac{S-2\pi .\frac{S}{6\pi }}{2\pi \sqrt{\frac{S}{6\pi }}}\Rightarrow h=\frac{3S-S}{3\times 2\pi }\times \sqrt{\frac{6\pi }{S}}$

$\Rightarrow h=\frac{2S}{6\pi }.\frac{\sqrt{6\pi }}{\sqrt{S}}=2\sqrt{\frac{S}{6\pi }}$

$\Rightarrow h=2r\left(diameter\right)$ $[\therefore r=\sqrt{\frac{S}{6\pi }}]$

Therefore, for maximum volume height of cylinder in equal to diameter of its base.