Question

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$

Answer 1

Let $R$ be the radius and $h$ be the height of cone.
$\therefore OA=h-r$
In $△OAB$
$r^2=R^2+(h-r{)}^2$
$\Rightarrow r^2=R^2+h^2+r^2-2rh$
$\Rightarrow R^2=2rh-h^2$
The volume $V$ of the cone is given by
$V=\frac{1}{3}\pi R^{2}h$
$=\frac{1}{3}\pi h(2rh-h^2)=\frac{1}{3}\pi (2r{h}^2-h^3)$
On differentiating with respect to $h$, we get
$\frac{dV}{dh}=\frac{1}{3}\pi (4rh-3h^2)$
For maximum and minimum ,put $\frac{dV}{dh}=0$
$\Rightarrow 4rh=3h^2$
$\Rightarrow 4r=3h$
$\Rightarrow h=\frac{4r}{3}...(h\ne 0)$
Now $\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi (4r-6h)$
At$h=\frac{4r}{3}$
${\left(\frac{d^{2}V}{d{h}^2}\right)}_{h=\frac{4r}{3}}=\frac{1}{3}\pi (4r-6\times \frac{4r}{3})$
$=\frac{\pi }{3}(4r-8r)$
$=\frac{-4r\pi }{3}<0$
$\Rightarrow V$ is maximum when $h=\frac{4r}{3}$
Hence, volume of the cone is maximum when $h=\frac{4r}{3}$, which is the altitude of cone.