A given right circular cone has a volume $p$ , and the largest right circular cylinder that can be inscribed in the cone has a volume $q.$ Then $\frac{8p}{9q}=$ is

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 $h=QL=OL-OQ=H-x\cot \alpha$ 

$V=$volume of the cylinder $=\pi x^2(H-x\cot \alpha )$

Also $p=\frac{1}{3}\pi {\left(H\tan \alpha \right)}^{2}H\mathrm{...............}\left(1\right)$

$\frac{dV}{dx}=\pi (2Hx-3x^2\cot \alpha )$ 

So, $\frac{dV}{dx}=0\iff x=0, x=\frac{2}{3}H\tan \alpha ,{\frac{d^{2}V}{d{x}^2}|}_{x=\frac{2}{3}H\tan \alpha }=-2\pi H<0.$

$V$ 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.$ [using (i)] 

Hence $p:q=9:4$