Question

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is (A) 6π (B) 3√3 π
(C) (4/3)π
(D) 2√3 π
[JEE Main 2019]

Answer 1

$\text{Let r, h and l be radius, height and slant height of the cone respecively}$

$\text{Then,}\;l^2=r^2+h^2$

$\implies\,9=r^2+h^2$

$\implies\,r^2=9-h^2$

$\text{Volume of the cone}$

$V=\frac{1}{3}\pi\,r^2\,h$

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

$V(h)=\frac{\pi}{3}(9h-h^3)$

$V'(h)=\frac{\pi}{3}(9-3h^2)$

$V''(h)=\frac{\pi}{3}(-6h)$

$\text{For maximum,}\;V'(h)=0$

$\frac{\pi}{3}(9-3h^2)=0$

$\implies\,9=3h^2$

$\implies\,h^2=3$

$\implies\,h=\pm\sqrt{3}$

$\text{But h cannot be negative}$

$\text{when h= \sqrt{3} ,}$

$V''(h)=\frac{\pi}{3}(-6\sqrt{3})<0$

$\therefore\text{V attains maximum at h=\sqrt{3} }$

$\text{Maximum volume of cone}$

$=V(\sqrt{3})$

$=\frac{\pi}{3}(9\sqrt{3}-3\sqrt{3})$

$=\frac{\pi}{3}(6\sqrt{3})$

$=\pi(2\sqrt{3})$

$=\bf\,2\sqrt{3}\,\pi\;\text{Cubic meter}$

$\therefore\textbf{Option (D) is correct}$