Question

Prove that the ratio of the height of the cone of maximum volume inscribed in a given sphere with the radius of the sphere is $4:3$.

Answer 1

The radius of the sphere is $r$ in which the cone is inscribed. Let $O$ be the centre of the sphere. Let $BC$ be the diameter of the base of the cone.

Let $AM$ be the altitude.

$\angle BOM=\theta$

Radius of the cone $R=rsin\theta$

Altitude of the cone $ABC,AM=AO+OM$

$\Rightarrow r+rcos\theta =r(1+cos\theta )$

We have volume of the cone $V=\frac{1}{3}\pi r^{2}h$

By substituting the value of $h$ we get

$V=\frac{1}{3}\pi r^{2}si{n}^2\theta \times r(1+cos\theta )$

$\therefore V=\frac{1}{3}\pi r^{3}si{n}^2\theta \times (1+cos\theta )$

Differentiate with respect to $\theta$ we get

$\frac{dV}{d\theta }=\frac{1}{3}\pi r^3\left[2sin\theta cos\theta \right(1+cos\theta )+si{n}^2\theta (-sin\theta \left)\right]$

$=\frac{1}{3}\pi r^{3}sin\theta [2cos\theta +2co{s}^2\theta -si{n}^2\theta ]$

$=\frac{1}{3}\pi r^{3}sin\theta [2cos\theta +2co{s}^2\theta -1+co{s}^2\theta ]$

$=\frac{1}{3}\pi r^{3}sin\theta [3co{s}^2\theta +2cos\theta -1]$

$\therefore \frac{dV}{d\theta }=\frac{1}{3}\pi r^{3}sin\theta (cos\theta +1)(3cos\theta -1)$

Thus $\frac{dV}{d\theta }=0$ at $cos\theta =\frac{1}{3}$

$cos\theta \ne -1$, since $\theta \ne \pi$

$\frac{dV}{d\theta }$ changes sign from positive to negative.

Hence $V$ is maximum at $cos\theta =\frac{1}{3}$

Altitude $=r(1+cos\theta )=r(1+\frac{1}{3})=\frac{4r}{3}$

Hence the ratio of the height of the cone of maximum volume inscribed in a given sphere with the radius of the sphere is $4:3$.