Question

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

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

Answer 1

Answer: (C)

$2 \sqrt{3} \pi$

Answer 2

$\therefore h = 3 \cos\theta$ $r=3 \sin\theta$ 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, $\frac{d^{2}V}{d\theta^{2}} \bigg]_{\sin\theta = \sqrt{\frac{2}{3}}} =$ 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)