Question

The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is

• A. $\frac{r}{2}$
• B. $\frac{r}{3}$
• C. $\frac{3r}{4}$
• D. $\frac{4r}{3}$

Answer 1

Answer: (D)

Let $R$ be the radius and $h$ be the height of cone.
$\therefore OA=h-r$
In $\Delta OAB$
$r^2=R^2+(h-r{)}^2$
$\Rightarrow r^2=R^2+h^2+r^2-2rh$
$\Rightarrow R^2=2rh-h^2$
The volume $V$ of the cone is given by
$V=\frac{1}{3}\pi R^{2}h$
$=\frac{1}{3}\pi h\left(2rh-h^2\right)=\frac{1}{3}\pi \left(2r{h}^2-h^3\right)$
On differentiating w.r.t. $h$, we get
$\frac{dV}{dh}=\frac{1}{3}\pi \left(4rh-3h^2\right)$
For maximum and minimum, put $\frac{dV}{dh}=0$
$\Rightarrow 4rh=3h^2$
$\Rightarrow 4r=3h$

$\Rightarrow h=\frac{4r}{3}$
Now, $\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi (4r-6h)$
At $h=\frac{4r}{3},$
${\left(\frac{d^{2}V}{d{h}^2}\right)}_{h=\frac{4r}{3}}=\frac{1}{3}\pi \left(4r-6\times \frac{4r}{3}\right)$
$=\frac{\pi }{3}(4r-8r)$
$=\frac{-4r\pi }{3}<0$
$\Rightarrow V$ is maximum when $h=\frac{4r}{3}.$
Hence, volume of the cone is maximum when $h=\frac{4r}{2}$, which is the altitude of cone.