Question

The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

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

Answer 1

Answer: (C)

Let the height of the cone $=h$
and the radius of the cone $=r$
Given, radius of the sphere $=R$
Now, In $\Delta OPB$

$\Rightarrow R^2=r^2+(h-R{)}^2$
$\Rightarrow r^2=R^2-(h-R{)}^2$
$=(R+h-R)(R-h+R)$
$\Rightarrow r^2=h(2R-h)$
The volume of the cone is
$V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow V=\frac{1}{3}\pi h(2R-h)h$
$\Rightarrow V=\frac{\pi }{3}\left(2R{h}^2-h^3\right)$
Differentiating with $r$ to $h$
$\frac{dV}{dh}=\frac{\pi }{3}\left(4Rh-3h^2\right)$
For maximum or minimum value of volume
$\frac{dV}{dh}=0$
$\Rightarrow \frac{\pi }{3}\left(4Rh-3h^2\right)=0$
$\Rightarrow h(4R-3h)=0$
$\Rightarrow h=0,h=\frac{4R}{3}$ (Not possible)
Now, $\frac{d^{2}V}{d{h}^2}=\frac{\pi }{3}(4R-6h)$
${\left(\frac{d^{2}V}{d{h}^2}\right)}_{\left(\text{at }h=\frac{4R}{3}\right)}=\frac{\pi }{3}\left(4R-6\cdot \frac{4R}{3}\right)$
$=\frac{\pi }{3}(4R-8R)=-\frac{4\pi }{3}R\Rightarrow$ Negative
ie., Maximum Hence, the height of the cone of maximum volume is
$\left(\frac{4R}{3}\right)$.