A given right cone has volume p, and the largest right circular cylinder that can be inscribed in the cone has volme q. Then $p : q$  is

Let $H$  be the height of the cone and $\alpha$  be its semi– vertical angle. Suppose that x is the raidius of the inscribed cylinder and h is its height. 

$h=QL=QL-OQ=H-x\cot \alpha$

$\Rightarrow \mathrm{V}=\mathrm{V}\text{olume of the cylinder}=\pi x^2(H-x\cot \alpha )$

Also.   $p=\frac{1}{3}\pi (H\tan \alpha {)}^{2}H\Rightarrow \frac{dV}{dx}=\pi (2Hx-3x^2\cot \alpha )$

So,  $\frac{dv}{dx}=0,\text{i.e .}x=0$  or  $x=\frac{2}{3}H t\mathrm{an}\alpha$

${\Rightarrow \frac{d^{2}V}{d{x}^2}|}_{x=\frac{2}{3}H\tan \alpha }=-2\pi H<0$

So, V is maximum when   $x=\frac{2}{3}H \tan \alpha$.

$\Rightarrow q=V_{max}=\pi \frac{4}{9}{H}^2{\tan }^2\alpha \frac{1}{3}H$

$=\frac{4}{27}\frac{{\pi }^{3}p{\tan }^2\alpha }{\pi {\tan }^2\alpha }=\frac{4}{9}p$  ; 

 Hence,  $p:q=9:4$