Question

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

Answer 1

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 $=\pi r^{2}h$

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

$=\pi \frac{x^4}{4}.h$

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

Substituting the value of $x^2$, we get

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

From (1), we have

$x^2=4R^2-h^2$

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

Differentiating with respect to $x$,

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

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

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

We know $\frac{dV}{h}=0$

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

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

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

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

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

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

$\frac{d^{2}V}{d{h}^2}=-\frac{3}{2}\pi \left(\frac{2R}{\sqrt{3}}\right)=-ve$

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

Maximum volume 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