Question

Show 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 h 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 = $\pi r^{2}h$

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

$=\pi \left(\frac{x^4}{4}\right)h$

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

Substitute 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 x,

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

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

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

$\frac{dV}{dh}=0$

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

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

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

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

$=-\frac{3}{2}.\pi h$

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

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

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

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

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

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

$=\frac{4\pi R^3}{3\sqrt{3}}$ Sq.units.