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 $\frac{p}{q}=$

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  $h=QL=OL-OQ=H-\cot \alpha$
V = volume of the cylinder  $=\pi x^2(H-x\cot \alpha )$
Also  $p=\frac{1}{3}\pi {\left(H\tan \alpha \right)}^{2}H\mathrm{.........}\left(i\right)$

$$\begin{gathered} \frac{dV}{dx}=\pi (2Hx-3x^2\cot \alpha ) \\ So, \frac{dV}{dx}=0\iff x=0,x=\frac{2}{3}H\tan \alpha ,{\frac{d^{2}V}{d{x}^2}|}_{x=\frac{2}{3}H\tan \alpha }=-2\pi H<0 \end{gathered}$$

So 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 $p:q = 9:4$