#### Question

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm ?

#### Solution

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

\[h = R + \sqrt{R^2 - r^2}\]

\[ \Rightarrow h - R = \sqrt{R^2 - r^2}\]

\[\text { Squaring both the sides, we get}\]

\[ h^2 + R^2 - 2hR = R^2 - r^2 \]

\[ \Rightarrow r^2 = 2hR - h^2 ........ \left( 1 \right)\]

\[\text { Now,} \]

\[V = \frac{1}{3}\pi r^2 h\]

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

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

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

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

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

\[ \Rightarrow 4hR = 3 h^2 \]

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

\[\text { Now,} \]

\[\frac{d^2 V}{d h^2} = \frac{\pi}{3}\left( 4R - 6h \right)\]

\[ \Rightarrow \frac{\pi}{3}\left( 4R - 8R \right) = 0\]

\[ \Rightarrow \frac{- 4\pi R}{3} < 0\]

\[\text { So, the volume is maximum when h } = \frac{4R}{3} . \]

\[ \Rightarrow h = \frac{4 \times 12}{3} = 16 cm\]