Question

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

Answer 1

Let VAB be the 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' is the height of the cylinder = VO - VO' = h - x cot $\alpha$
Volume of the cylinder,
$V=\pi x^2(h-xcot\alpha )$ ...(i)
On differentiating w.r.t.x, we get
$\frac{dV}{dx}=2\pi hx-3\pi x^{2}cot\alpha$
For maxima or minima put $\frac{dV}{dx}=0$
$\Rightarrow 2\pi xh-3\pi x^{2}cot\alpha =0$
$\Rightarrow x=\frac{2h}{3}tan\alpha (\because x\ne 0$
Now, $\frac{d^{2}V}{d{x}^2}=2\pi h-6\pi xcot\alpha$
At $x=\frac{2h}{3}tan\alpha ,$ $\frac{d^{2}V}{d{x}^2}=\pi (2h-4h)=-2\pi h<0$
$\Rightarrow V$ Maximum, when $x=\frac{2h}{3}tan\alpha$. Now OO' $=h-xcot\alpha -h-\frac{2h}{3}=\frac{h}{3}$
$\therefore$ The maximum volume of the cylinder is
$V=\pi \left(\frac{2h}{3}tan\alpha {)}^2\right(h-\frac{2h}{3})=\frac{4}{27}\pi h^{3}ta{n}^2\alpha$