Question

Find the volume of the largest cone can be inscribed in a sphere of radius R.

Answer 1

Let $r$ be the base radius $x$ is the distance $O$ the center of the sphere from the base and $V$ the volume of the are

Height $h$ of the cone $=R+x$

$\therefore$ $V=\frac{1}{3}\pi r^{2}h=\frac{\pi }{3}(R^2-x^2)(R+x)$

$=\frac{\pi }{3}(R^2+R^{2}x-R{x}^2-x^2)$

$\therefore$ $\frac{dV}{dx}=\frac{\pi }{3}[R^2-2Rx-{3x}^2]$

$\frac{d^{2}V}{{dx}^2}=\frac{\pi }{3}[-2R-6x]$

For max or min $V\frac{dV}{dx}=0$

$\therefore$ $R^2-2Rx-{3x}^2=0$

$\Rightarrow (R+x)(x-3x)=0$ 2) $x=-R,\frac{x}{3}$ but $x\ne -R$

When $x=\frac{R}{3}\frac{d^{2}V}{{dx}^2}<0$ $V$ is max only when $x=\frac{R}{3}$

$\therefore$ Max $V=\frac{1}{3}\pi (R^2-\frac{R^2}{9})(R+\frac{R}{3})=\frac{{32\pi R}^3}{81}=\frac{8}{27}\left(\frac{4}{3}\pi R^3\right)=\frac{8}{27}$ (volume of sphere)