Question

The right circular cone of largest volume that can be enclosed by a sphere of $1m$ radius has height of

• A. $\frac{1}{3}m$
• B. $\frac{2}{3}m$
• C. $\frac{2\sqrt{2}}{3}m$
• D. $\frac{4}{3}m$

Answer 1

The correct option is D $\frac{4}{3}m$

Let height (AM) of the cone is $h$ meters.


OA=OB=1m
$\therefore$ Radius of base of right circular cone is
$r=BM=(O{B}^2-O{M}^2{)}^{1/2}$
$r=(1-(h-1{)}^2{)}^{1/2}$
$r=(2h-h^2{)}^{1/2}$
$\therefore$ Volume of the cone
$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (2h-h^2)h$
For maximum volume $V$
$\frac{dV}{dh}=0\Rightarrow \frac{\pi }{3}(4h-3h^2)=0$
$\Rightarrow h=0orh=\frac{4}{3}$
But $h=0$ (Not possible)
So, required height, $h=\frac{4}{3}m$