Question

Prove that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$. Also find the maximum volume.

Answer 1

Given:

Radius of the sphere $R.$

Let $x$ be the diameter of the base of the inscribed cylinder .

Then,

$h^2+x^2=(2R{)}^2$

$h^2+x^2=4R^2$---(1)

Volume of the cylinder =$\pi r^{2}h$

$V=\pi {\left(\frac{x}{2}\right)}^2.h$

$V=\frac{1}{4}\pi x^{2}h$

Substituting the value of $x^2$ we get

$V=\frac{1}{4}\pi h(4R^2-h^2)$

From (1), $x^2=4R^2-h^2$

$V=\pi R^{2}h-\frac{1}{4}\pi h^3$

Differentiating with respect to $h$,

$\frac{dV}{dh}=\pi R^2-\frac{3}{4}\pi h^2$

$=\pi [R^2-\frac{3}{4}{h}^2]$

For Max volume, $\frac{dV}{dh}=0$

$\pi [R^2-\frac{3}{4}{h}^2]=0$

$\Rightarrow R^2=\frac{3}{4}{h}^2$

$\Rightarrow h=\frac{2R}{\sqrt{3}}$

Also, $\frac{d^{2}V}{d{h}^2}=-\frac{3}{4}.2\pi h$ $=\frac{-3}{2}\pi h$

At $h=\frac{2R}{\sqrt{3}}$

$\frac{d^{2}V}{d{h}^2}=-\frac{3}{2}.\pi \frac{2R}{\sqrt{3}}<0$

$V$ is maximum at $h=\frac{2R}{\sqrt{3}}$

Maximum volume at $h=\frac{2R}{\sqrt{3}}$

$\Rightarrow V_{max}=\frac{1}{4}\pi \left[\frac{2R}{\sqrt{3}}\right][4R^2-\frac{4R^2}{3}]$

$\Rightarrow V_{max}=\frac{\pi R}{2\sqrt{3}}\left[\frac{8R^2}{3}\right]$

$\Rightarrow V_{max}=\frac{4\pi R^3}{3\sqrt{3}}$ sq.units.