Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

(N/A) Let $r$ and $h$ be the radius and height of the cone respectively inscribed in a sphere of radius $R$.
Let $V$ be the volume of the cone.
Then,$V = \frac{1}{3} \pi r^{2} h$.
Height of the cone is given by $h = R + AB = R + \sqrt{R^{2} - r^{2}}$.
$V = \frac{1}{3} \pi r^{2} (R + \sqrt{R^{2} - r^{2}})$.
$V = \frac{1}{3} \pi r^{2} R + \frac{1}{3} \pi r^{2} \sqrt{R^{2} - r^{2}}$.
Differentiating with respect to $r$:
$\frac{dV}{dr} = \frac{2}{3} \pi r R + \frac{1}{3} \pi \left( 2r \sqrt{R^{2} - r^{2}} + r^{2} \cdot \frac{-2r}{2\sqrt{R^{2} - r^{2}}} \right)$.
$\frac{dV}{dr} = \frac{2}{3} \pi r R + \frac{1}{3} \pi \left( \frac{2r(R^{2} - r^{2}) - r^{3}}{\sqrt{R^{2} - r^{2}}} \right) = \frac{2}{3} \pi r R + \frac{\pi r (2R^{2} - 3r^{2})}{3\sqrt{R^{2} - r^{2}}}$.
Setting $\frac{dV}{dr} = 0$:
$\frac{2}{3} \pi r R = - \frac{\pi r (2R^{2} - 3r^{2})}{3\sqrt{R^{2} - r^{2}}} \Rightarrow 2R\sqrt{R^{2} - r^{2}} = 3r^{2} - 2R^{2}$.
Squaring both sides: $4R^{2}(R^{2} - r^{2}) = (3r^{2} - 2R^{2})^{2}$.
$4R^{4} - 4R^{2}r^{2} = 9r^{4} + 4R^{4} - 12r^{2}R^{2}$.
$9r^{4} = 8R^{2}r^{2} \Rightarrow r^{2} = \frac{8}{9}R^{2}$.
Then $h = R + \sqrt{R^{2} - \frac{8}{9}R^{2}} = R + \frac{R}{3} = \frac{4}{3}R$.
Maximum volume $V = \frac{1}{3} \pi (\frac{8}{9}R^{2}) (\frac{4}{3}R) = \frac{32}{81} \pi R^{3}$.
Volume of sphere $V_{s} = \frac{4}{3} \pi R^{3}$.
Ratio $\frac{V}{V_{s}} = \frac{32/81 \pi R^{3}}{4/3 \pi R^{3}} = \frac{32}{81} \cdot \frac{3}{4} = \frac{8}{27}$.
Hence,the volume of the largest cone is $\frac{8}{27}$ of the volume of the sphere.