Question

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\frac{4r}{3}$. Also show that the maximum volume of the cone is $\frac{8}{27}$ of the volume of the sphere.

Answer 1

Radius of sphere=r

Radius of cone=R

Height of cone=h

$BC=\sqrt{r^2-R^2}$

$\therefore h=r+\sqrt{r^2-R^2}$

$\therefore$Volume of cone$=\frac{1}{3}\pi R^2(r+\sqrt{r^2-R^2})$

$\frac{dV}{dR}=\frac{2}{3}\pi Rr+\frac{2\pi }{3}R\sqrt{r^2-R^2}+\frac{\pi R^2}{3}.\frac{(-2R)}{2\sqrt{r^2-R^2}}$

$=\frac{2\pi Rr}{3}+\frac{2\pi R(r^2-R^2)-\pi R^3}{3\sqrt{r^2-R^2}}$

$=\frac{2\pi Rr}{3}+\frac{2\pi R{r}^2-3\pi R^3}{3\sqrt{r^2-R^2}}$

For max. volume

$\frac{dV}{dR}=0$

$\Rightarrow \frac{2\pi Rr}{3}=\frac{3\pi R^2-2\pi r^2}{3\sqrt{r^2-R^2}}$

$\Rightarrow 2Rr=\frac{3R^3-2r^{2}R}{\sqrt{r^2-R^2}}\Rightarrow 2r\sqrt{r^2-R^2}=3R^2-2r^2$

$\Rightarrow 9R^4-8r^2{R}^2=0$

$\therefore R^2=0$

$\frac{8r^2}{9}$

$\frac{d^{2}V}{d{R}^2}<0$

$R^2=\frac{8r^2}{9}$

$\therefore$For $V_{max},R^2=\frac{8r^2}{9}$

$\therefore h=r+\frac{r}{3}=\frac{4r}{3}$

$V=\frac{1}{3}\times \frac{8r^2}{9}\times \frac{4r}{3}=\frac{8}{27}\frac{4\pi r^3}{3}$