Question

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

A cone of maximum volume is inscribed in a given sphere. Then prove that ratio of the height of the cone to the diameter of the sphere is equal to `2/3`.

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\text{Now,}\]

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

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

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

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

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

\[\text{Hence proved}.\]