Question

The volume of the greatest cylinder which can be inscribed in a cone of height $h$ and semi-vertical angle $\alpha$ is (in cubic units)

• A. $\frac{4}{27}\pi h^{3}\,tan^{2}\,\alpha$
• B. $\frac{4}{27}\pi h^{2}\,tan^{2}\,\alpha$
• C. $\pi h^{3}\,tan^{2}\,\alpha$
• D. $\frac{5}{36}\pi h\,tan\,\alpha$

Answer 1

Answer: (A)

Let $VAB$ be a given cone of height $h$, semi-vertical angle $\alpha$ and let $x$ be the radius of the base of the cylinder $A'B'DC$ which is inscribed in the cone $VAB$. Then,
$OO' =$ Height of the cylinder
$\Rightarrow OO' -VO - VO' = h - x \,cot \,\alpha\quad\ldots\left(i\right)$
Let $V$ be the volume of the cylinder. Then,
$V = \pi x^{2}\left( h - x \,cot \,\alpha \right)\quad \ldots \left(ii\right)$
$\Rightarrow \frac{dV}{dx} = 2\pi\,xh -3\pi\, x^2 \,cot \,\alpha $
For maximum or minimum $V$, we must have
$\frac{dV}{dx} =0$
$\Rightarrow 2\pi xh -3\pi x^{2} \,cot \,\alpha = 0$



$\Rightarrow x = \frac{2h}{3}tan\,\alpha \left[\because x \ne 0\right]$
Now, $\frac{d^{2}V}{dx^{2}} = 2\pi h -6\pi \, x \,cot \,\alpha $
When $x = \frac{2h}{3}tan\,\alpha$, we have
$\frac{d^{2}V}{dx^{2}} =\pi\left[2h-4h\right] = 2\pi h < 0$.
Hence, $V$ is maximum when $x = \frac{2h}{3}tan\,\alpha$.
The maximum volume of the cylinder is given by
$V = \pi\left(\frac{2h}{3}tan\,\alpha \right)^{2}\left(h -\frac{2h}{3}\right)$ [From $\left(ii\right)$]
$\Rightarrow V = \frac{4}{27}\pi h^{3}\,tan^{2}\,\alpha $.