Question

A cone of maximum volume is inscribed in a given sphere. Find the ratio of the height of the cone to the diameter of the sphere.

Answer 1

In $△AOB,{AO}^2+{AB}^2={OB}^2$

$\Rightarrow {(h-R)}^2+r^2=R^2$

$\Rightarrow r^2=R^2-{(h-R)}^2$ ......$\left(1\right)$

Volume of cone$=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi [R^2-{(h-R)}^2]h$ .....$\left(2\right)$

For maximum value of $V$

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

$\Rightarrow \frac{dV}{dh}=\frac{1}{3}\pi [-2(h-R)h+R^2-{(h-R)}^2]=0$

$\Rightarrow -2(h-R)h+R^2-{(h-R)}^2=0$

$\Rightarrow -2h^2+2Rh+R^2-h^2+2hR-R^2=0$

$\Rightarrow -3h^2+4hR=0$

$\Rightarrow -3h+4R=0$

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

$\therefore \frac{h}{2R}=\frac{4}{6}=\frac{2}{3}$

Thus, the ratio of the height of cone to the diameter of the sphere for maximum volume of the cone $=\frac{2}{3}$