Question

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 $p:q$ is

• A. 9: 4
• B. 8: 3
• C. 7: 2
• D. none of these

Answer 1

The correct option is A 9: 4

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

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

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

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

So $\frac{dV}{dx}=0\iff x=0,x=\frac{2}{3}H\tan \alpha$,

At $x=\frac{2}{3}H\tan \alpha$

$\frac{d^{2}V}{{dx}^2}=-2\pi H<0$

So $V$ is maximum when

$x=\frac{2}{3}H\tan \alpha$

and $q=V_{max}=\pi \frac{4}{9}{H}^2{\tan }^2\alpha \frac{1}{3}H=\frac{4}{27}\frac{\pi 3p{\tan }^2\alpha }{\pi {\tan }^2\alpha }=\frac{4}{9}p.$

Hence $p:q=9:4$