Question

Which of the following is/are true for the right circular cone with maximum volume (in cu.m) and having slant height $3m$

• A.
  $h=\sqrt{6}$
• B. $h=\sqrt{3}$
  
• C. $V_{max}=2\sqrt{3}\pi$
• D. $V_{max}=\sqrt{6}\pi$
  

Answer 1

Answer: (B), (C)

$V_{max}=2\sqrt{3}\pi$

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,
$V=\frac{1}{3}\pi (9-h^2)h=\frac{1}{3}\pi (9-(\sqrt{3}{)}^2)\sqrt{3}=2\sqrt{3}\pi m^3$