Question

The volume of the greatest cone obtained by rotating a right-angled triangle of hypotenuse $2$ units about a side is $\frac{k\pi }{9\sqrt{3}}$ cubic units, then the value of $k$ is equal to

Answer 1

Answer: 16

The volume generated $=V=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}\pi \left(4-x^2\right)\left(x\right)$
For maximum volume, $\frac{dV}{dx}=0$
$\Rightarrow 4-3x^2=0\Rightarrow x=\frac{2}{\sqrt{3}}$
$V_{max}=\frac{\pi }{3}\left(4-\frac{4}{3}\right)\frac{2}{\sqrt{3}}$ $=\frac{16\pi }{9\sqrt{3}}$
Also, $\frac{d^{2}V}{d{x}^2}<0$