Explanation

(a)
Let $H$ be the height of the cone and $\alpha$ be its semi-vertical angle. Suppose that $x$ is the radius of the inscribed cylinder and $h$ be its height
$\therefore\ \ h = QL = OL - OQ$
$= H - x\cot\alpha$
$V =$Volume of the cylinder
$= \pi x^{2}(H - x\cot\alpha)$
Also, $p = \frac{1}{3}\pi\left( H\tan\alpha \right)^{2}H$ …(i)
$\frac{dV}{dx} = \pi(2Hx - 3x^{2}\cot\alpha)$
$\therefore\ \frac{dV}{dx} = 0\ \ \ \Rightarrow \ \ \ \ x = 0$,
$x = \frac{2}{3}H\tan\alpha$
$\Rightarrow \ \ \left. \ \frac{d^{2}V}{dx^{2}} \right|_{x = \frac{2}{3}H\tan\alpha} = - 2\pi\ H < 0$
[image:6_image16.emf]
$\therefore\ \ V$ is maximum when $x = \frac{2}{3}H\tan\alpha$
And $q = V_{\max} = \pi\frac{4}{9}H^{2}\tan^{2}\alpha\frac{1}{3}H$
$= \frac{4}{9}p$ [from Eq. (i)]
Hence, $p:q = 9:4$