#### Question

Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to \[ \frac{2}{3} \] of the diameter of the sphere.

#### Solution

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

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

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

\[ \Rightarrow h^2 + R^2 - 2hr = R^2 - r^2 \]

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

\[\text { Volume of cone } = \frac{1}{3}\pi r^2 h\]

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

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

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

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

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

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

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

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

\[\text { Substituting the value of y in eq } . \left( 1 \right), \text { we get }\]

\[ x^2 = 4\left( r^2 - \left( \frac{r}{\sqrt{2}} \right)^2 \right)\]

\[ \Rightarrow x^2 = 4\left( r^2 - \frac{r^2}{2} \right)\]

\[ \Rightarrow x^2 = 4\left( \frac{r^2}{2} \right)\]

\[ \Rightarrow x^2 = 2 r^2 \]

\[ \Rightarrow x = r\sqrt{2}\]

\[\text { Now,} \]

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

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

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

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

\[ \Rightarrow h = \frac{2}{3}\left( \text { Diameter of sphere } \right)\]

\[\text { Hence proved }.\]