Question

The ratio of the altitude of the cone of greatest volume which can be inscribed in a given sphere, to the diameter of the sphere is

• A. 2/3
• B. 3/4
• C. 1/3
• D. 1/4

Answer 1

The correct option is A 2/3

Let $AD=x$ be the height of the cone $ABC$ inscribed in a sphere of radius a

$\therefore OD=x-a$

Then radius of its base

$r=CD=\sqrt{{\left(OC\right)}^2+{\left(OD\right)}^2}=\sqrt{a^2{(x-a)}^2}=\sqrt{2ax-x^2}$

Therefore volume $V$ of the cone is given by

$V=\frac{1}{3}\pi r^{2}x=\frac{1}{3}\pi ({2ax}^2-x^2)=\frac{1}{3}\pi ({2ax}^2-x^2)$

$\frac{dV}{dx}=\frac{1}{3}\pi (4ax-{3x}^2)$

$\frac{d^{2}V}{{dx}^2}=\frac{1}{3}\pi (4a-6x)$

For max or min of $V$

$\frac{dV}{dx}=0\Rightarrow x=\frac{4a}{3}$

For this value of $V$

$\frac{d^{2}V}{{dx}^2}=-\left(\frac{4\pi a}{3}\right)$ -ve

Therefore $V$ is max when $x=\frac{4a}{3}=\left(\frac{2}{3}\right)\left(2a\right)$

i.e when the height of cone $=x=\left(\frac{2}{3}\right)$ of the diameter of sphere