Question

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

Answer 1

$\text{Let 'a' be the radius of cylinder}\text{and 2x be the height of the cylinder then,}$
$x^2+a^2=R^2....\left(i\right)$

$\text{Let V be the volume of the cylinder then,}$

$V=\pi a^2.2x$

$=\pi (R^2-x^2)2x$

$=2\pi (R^{2}x-x^3),0<x<R$

$\frac{dV}{dx}=2\pi (R^2-3x^2)$

$\frac{d^{2}V}{d{x}^2}=2\pi (-6x)$,
$\text{For maxima a minima}$ $\frac{dV}{dx}=0$

$\Rightarrow R^2-3x^2=0\Rightarrow x^2=\frac{R^2}{3}\Rightarrow x=\frac{R}{\sqrt{3}}$

$Atx=\frac{R}{\sqrt{3}},\frac{d^{2}V}{d{x}^2}<0$
$\therefore Atx=\frac{R}{\sqrt{3}}$ $\text{cylinder will have the maximum volume.}$

$\therefore$ $\text{Height of the cylinder of maximum volume}$
$2x=\frac{2R}{\sqrt{3}}$
$\text{Maximum volume =}$$2\pi (R^2.\frac{R}{\sqrt{3}}-\frac{R^3}{3\sqrt{3}})$
$=\frac{2\pi }{\sqrt{3}}{R}^3(1-\frac{1}{3})=\frac{2\pi R^3}{\sqrt{3}}\left(\frac{2}{3}\right)$
$=\frac{4\pi R^3}{3\sqrt{3}}$