Let a cone. VAB of greatest volume be inscribed in the sphere let $\mathrm{AOC}=\theta$
$\therefore$ AC, radius of the base of the cone $=\mathrm{R} \sin \theta$


and $\mathrm{VC}=\mathrm{VO}+\mathrm{OC}$
$=\mathrm{R}(1+\cos \theta)$
$=\mathrm{R}+\mathrm{R} \cos \theta$
$=$ height of the cone.,
$\mathrm{V}$, the volume of the cone.
$\mathrm{V}=\frac{1}{3} \pi(\mathrm{AC})^2(\mathrm{VC})$
$\Rightarrow \mathrm{V}=\frac{1}{3} \pi \mathrm{R}^3 \sin ^2 \theta(1+\cos \theta)$
$\therefore \frac{\mathrm{dV}}{\mathrm{d} \theta}=\frac{1}{3} \pi \mathrm{R}^3\left(-3 \sin ^3 \theta+2 \sin \theta+2 \sin \theta \cos \theta\right)$
For maximum and minimum, we have $\frac{\mathrm{dV}}{\mathrm{d} \theta}=0 \Rightarrow \cos \theta=\frac{1}{3}$ or $\cos \theta=-1$ But $\cos \theta \neq-1$ as $\cos \theta=-1 \Rightarrow \theta=\pi$, which is not possible $\quad \therefore \cos \theta=\frac{1}{3}$
When $\cos \theta=\frac{1}{3}, \sin \theta=\sqrt{1-\cos ^2 \theta}=\frac{2 \sqrt{2}}{3}$ $\left(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \theta^2}\right)$ at $\theta=\cos ^{-1}\left(\frac{1}{3}\right) < 0$
Hence $\mathrm{V}$ is maximum at $\theta=\cos ^{-1}\left(\frac{1}{3}\right)$
Now, $\cos \theta=\frac{1}{3}, \sin \theta=\frac{2 \sqrt{2}}{3}$
$\therefore$ Maximum volume of cone. $=\frac{8}{27}\left(\frac{4}{3} \pi \mathrm{R}^3\right)$
Max. volume $=\frac{8}{27} \times$ volume of the sphere of cone.