Question

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

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

Answer 1

Answer: (C)

$\therefore h=3\cos \theta$
$r=3\sin \theta$
Now,
$V=\frac{1}{3}\pi r^{2}h=\frac{\pi }{3}\left(9{\sin }^2\theta \right).\left(3\cos \theta \right)$
$\therefore \frac{dV}{d\theta }=0\Rightarrow \sin \theta =\sqrt{\frac{2}{3}}$
Also, $\frac{d^{2}V}{d{\theta }^2}{]}_{\sin \theta =\sqrt{\frac{2}{3}}}=$ negative
$\Rightarrow$ Volume is maximum,
when $\sin \theta =\sqrt{\frac{2}{3}}$
$\therefore V_{max}\left(\sin \theta =\sqrt{\frac{2}{3}}\right)=2\sqrt{3}\pi$ (in cu. m)