Question

Show that the altitude of the right circular cone of maximum volume that can be iscribed 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 $\Delta OAB,r^2=R^2+(h-r{)}^2\Rightarrow r^2+h^2+r^2-2rh\Rightarrow R^2=2rh-h^2$
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 w.r.t.h, we get
$\frac{dV}{dh}=\frac{1}{3}\pi (4rh-3h^2)$
For maximum or minimum put $\frac{dV}{dh}=0$
$\Rightarrow 4rh=3h^2\Rightarrow 4r=3h$
$\therefore h=\frac{4r}{3}$ $(h\ne 0)$
Now, $\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi (4r-6h)$
$Ath=\frac{4r}{3},(\frac{d^{2}V}{d{h}^2}{)}_{h=\frac{4r}{3}}=\frac{1}{3}\pi (4r-6\times \frac{4r}{3})=\frac{\pi }{3}(4r-8r)=\frac{-4r\pi }{3}<0$
$\Rightarrow$ 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.