Question

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:

• A. $6\pi$
• B. $3\sqrt{3}\pi$
• C. $2\sqrt{3}\pi$
• D. $\frac{4}{3}\pi$

Answer 1

The correct option is C $2\sqrt{3}\pi$

Given :
Slant height of the cone, $L=3m.$
Let the radius be '$r$' and height be '$h$'.

Volume of a right circular cone,
$V=\frac{1}{3}\pi r^{2}h$
$\Rightarrow V=\frac{1}{3}\pi (L^2-h^2)h[\because L^2=r^2+h^2]\Rightarrow V=\frac{1}{3}\pi (9-h^2)h$

Differentiating $V$ w.r.t. $h$ and equating to $0$, we get
$\frac{dV}{dh}=\frac{1}{3}\pi (9-3h^2)=0\Rightarrow h=\pm \sqrt{3}$
As height cannot be negative so, $h=\sqrt{3}m$
Again differentiating $V$ w.r.t. $h$,
$\frac{d^{2}V}{d{h}^2}=-2\pi h<0$ at $h=\sqrt{3}m$
$\therefore$ Maximum volume occurs at $h=\sqrt{3}$.

Hence, the maximum volume of the cone,
$V=\frac{1}{3}\pi (9-h^2)h=\frac{1}{3}\pi (9-(\sqrt{3}{)}^2)\sqrt{3}=2\sqrt{3}\pi m^3$