#### Question

Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius \[5\sqrt{3 cm} \text { is }500 \pi {cm}^3 .\]

#### Solution

\[\text { Let the height, radius of base and volume of a cylinder be h, r and V, respectively . Then }, \]

\[\frac{h^2}{4} + r^2 = R^2 \]

\[ \Rightarrow h^2 = 4\left( R^2 - r^2 \right)\]

\[ \Rightarrow r^2 = R^2 - \frac{h^2}{4} . . . \left( 1 \right)\]

\[\text { Now }, \]

\[V = \pi r^2 h\]

\[ \Rightarrow V = \pi\left( h R^2 - \frac{h^3}{4} \right) \left[\text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dV}{dh} = \pi\left( R^2 - \frac{3 h^2}{4} \right)\]

\[\text { For maximum or minimum values of V, we must have }\]

\[\frac{dV}{dh} = 0\]

\[ \Rightarrow \pi\left( R^2 - \frac{3 h^2}{4} \right) = 0\]

\[ \Rightarrow R^2 - \frac{3 h^2}{4} = 0\]

\[ \Rightarrow R^2 = \frac{3 h^2}{4}\]

\[ \Rightarrow h = \frac{2R}{\sqrt{3}}\]

\[\frac{d^2 V}{d h^2} = \frac{- 3\pi h}{2}\]

\[\frac{d^2 V}{d h^2} = \frac{- 3\pi}{2} \times \frac{2R}{\sqrt{3}}\]

\[ \Rightarrow \frac{d^2 V}{d h^2} = \frac{- 3\pi R}{\sqrt{3}} < 0\]

\[\text { So, the volume is maximum when h } = \frac{2R}{\sqrt{3}} . \]

\[\text { Maximum volume } = \pi h\left( R^2 - \frac{h^2}{4} \right)\]

\[ = \pi \times \frac{2R}{\sqrt{3}}\left( R^2 - \frac{4 R^2}{12} \right)\]

\[ = \frac{2\pi R}{\sqrt{3}}\frac{8 R^2}{12}\]

\[ = \frac{4\pi R^3}{3\sqrt{3}}\]

\[ = \frac{4\pi \left( 5\sqrt{3} \right)^3}{3\sqrt{3}}\]

\[ = 500\pi {cm}^3\]