Question

A cone is inscribed in a sphere of radius $12$cm. If the volume of the cone is maximum, find its height.

Answer 1

A cone is inscribed in a sphere of radius $12cm$. We need to find its height if the volume of the cone is maximum.

Let a cone of radius $rcm$ and height $hcm$ is inscribed in a sphere of radius $12cm$.

In right triangle $OAB,(12{)}^2=(h-12{)}^2+r^2$

$\Rightarrow r^2=24h-h^2$

We know that Volume $V=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi (24h-h^2)h$

$\Rightarrow V=\frac{1}{3}\pi (24h^2-h^3)$

$\Rightarrow \frac{dV}{dh}=\frac{1}{3}\pi (48h-3h^2)$

For maximum volume: $\frac{dV}{dh}=0$

$\therefore \frac{1}{3}\pi (48h-3h^2)=0$

$\Rightarrow 48h-3h^2=0$

$\Rightarrow h=16$

Also $\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi (48-6h)$

${\left(\frac{d^{2}V}{d{h}^2}\right)}_{h=16}=\frac{1}{3}\pi (48-96)<0$

Therefore for $h=16$ the volume is maximum.