Question

The volume of the greatest cylinder which can be inscribed in a cone of height $30cm$ and semi-vertical angle ${30}^0$ is

• A. $\frac{4000\pi }{3}c{m}^3$
• B. $\frac{400\pi }{3}c{m}^3$
• C. $\frac{4000\pi }{\sqrt{3}}c{m}^3$
• D. None of the above.

Answer 1

The correct option is A $\frac{4000\pi }{3}c{m}^3$

Let VAB be a given coe of heights h,semi-vertical angle $\alpha$ and let $x$ be the radius of the base of the cylinder ABDC
that is increased in cone VAB
then OO'=height of the cylinder ,OO'=VO-VO'=$h-x\cot \alpha$
Let V be the volumn of the cylinder then.
$v=\pi x^2(h-x\cot \alpha )$
$\frac{dv}{dx}=2\pi xh-3\pi x^2\cot \alpha$

For maximumm and minimum V we must have,

$\frac{dv}{dx}=0=2\pi xh-3\pi x^2\cot \alpha =0$

$\Rightarrow x=\frac{2h}{3}\tan \alpha (\because x\ne 0)$

$\frac{d^{2}v}{d{x}^2}=2\pi h-6\pi x\cot \alpha henx=\frac{2h}{3}\tan \alpha$

we have $\frac{d^{2}v}{d{x}^2}=\pi [2h-4h]=-2h<0$
Hence V is maximum when $x=\frac{2h}{3}\tan \alpha$
So the maximum value of the cylinder is given by
$v={\left(\frac{2h}{3}\tan \alpha \right)}^2(h-\frac{2h}{3})$
$v=\frac{4}{27}\pi h^3{\tan }^2\alpha$
At $h=h=3\alpha m$

$\alpha =30°$
$v=\frac{4}{27}\pi {30}^3{\tan }^{2}30°$
$\frac{4\pi \times 27\times 1000}{27}{\left(\frac{1}{\sqrt{3}}\right)}^2$
$v=\frac{4000\pi }{3}{com}^3$
Hence the correct option is A