JEE PYQ: Application Of Derivatives Question 69
Question 69 - 2019 (09 Jan Shift 1)
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:
(1) $6\pi$
(2) $3\sqrt{3},\pi$
(3) $\frac{4}{3}\pi$
(4) $2\sqrt{3},\pi$
Show Answer
Answer: (4)
Solution
$h^2 + r^2 = 9$. $V = \frac{1}{3}\pi r^2 h = \frac{\pi}{3}(9-h^2)h$. $\frac{dV}{dh} = \frac{\pi}{3}(9-3h^2) = 0 \Rightarrow h = \sqrt{3}$. $V = \frac{\pi}{3}(9-3)\sqrt{3} = 2\sqrt{3}\pi$.
BETA
