Question

Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle $\alpha$ is $\frac{4}{27}\pi h^3{\tan }^2\alpha$.

Answer 1

Consider $VAB$ be a given cone of height $h$, semi vertical angle $\alpha$

And let $x$ be the radius of base of the cylinder $A$ $B$ $DC$ which is inscribed in the cone $VAB$

$oo=heightofthecylinder$

$\Rightarrow oo=VO-VO=h-x\cot \alpha$

Let $V$ be the volume of the cylinder

Then,

$V=\pi x^2\left(h-x\cot \alpha \right).........\left(1\right)$

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

For maximum of minimum $V$ we must have

$\begin{array}{c}\frac{dV}{dx}=0\\ \Rightarrow 2\pi xh-3\pi x^2\cot \alpha =0\\ \Rightarrow x=\frac{2h}{3}\tan \alpha \left(\because x\ne 0\right)\end{array}$

Now, $\frac{d^{2}V}{d{x}^2}=2\pi h-6\pi x\cot \alpha$

When, $x=\frac{2h}{3}\tan \alpha$ we have

$\begin{array}{c}\frac{d^{2}V}{d{x}^2}=\pi \left[2h-4h\right]\\ =-2\pi h<0\end{array}$

Hence, $V$ is maximum when $x=\frac{2h}{3}\tan \alpha$

And the maximum volume of cylinder is

$V=\pi {\left(\frac{2h}{3}\tan \alpha \right)}^2\left(h-\frac{2h}{3}\right)\left(from\left(1\right)\right)$

$\Rightarrow V=\frac{4}{27}\pi h^3{\tan }^2\alpha$