Question

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

Answer 1

The correct answer is $\frac{4r}{3}.$
A sphere of fixed radius $(r)$ is given.
Let $R$ and $h$ be the radius and the height of the cone respectively.

The volume $(V)$ of the cone is given by,
$V=\frac{1}{3}πR^2h$
Now, from the right triangle BCD, we have:
$BC=\sqrt{r^2-R^2}$
$∴h=r+\sqrt{r^2-R^2}$
∴V=\frac{1}{3}πR^2(r+\sqrt{r^2-R^2})$$=\frac{1}{3}πR^2r+\frac{1}{3}πR^2\sqrt{r^2-R^2}
$∴\frac{dV}{dR}=\frac{2}{3}πRr+\frac{2π}{3}πR\sqrt{r^2-R^2}+\frac{R^2}{3}.\frac{(-2R)}{2\sqrt{r^2-R^2}}$
$=\frac{2}{3}πRr+\frac{2π}{3πR}\sqrt{r^2-R^2}-\frac{R^3}{3\sqrt{r^2-R^2}}$
$=\frac{2}{3}πRr+\frac{2πR(r^2-R^2)-\pi R^3}{3\sqrt{r^2-R^2}}$
$=\frac{2}{3}πRr+\frac{2πRr^2-3\pi R^3}{3\sqrt{r^2-R^2}}$
Now,$\frac{dV}{dR^2}=0$
$⇒\frac{2πrR}{3}=\frac{3πR^3-2πRr^3}{3\sqrt{r^2-R^2}}$
$⇒2r\sqrt{r^2-R^2}=3R^2-2r^2$
$⇒4r^2(r^2-R^2)=3R^2-2r^2$
$=4r^4-4r^2R^2=9R^4+4r^4-12R^2r^2$
$⇒9R^4-8r^2R^2=0$
$⇒9R^2=8r^2$
$⇒R^2=\frac{8r^2}{9}$
Now,$\frac{d^2V}{dR^2}=\frac{2πr}{3}+\frac{3\sqrt{r^2-R^2}(2πr^2-9πR^2)-(2πRr^2-3πR^3)(-6R)\frac{1}{2\sqrt{r^2-R^2}}}{9(r^2-R^2)}$
$=\frac{2πr}{3}+\frac{3\sqrt{r^2-R^2}(2πr^2-9πR^2)-(2πRr^2-3πR^3)(3R)\frac{1}{2\sqrt{r^2-R^2}}}{9(r^2-R^2)}$
Now when $R^2=\frac{8r^2}{9}$,it can be shown that $\frac{d^2V}{dR^2}<0.$
∴ The volume is the maximum when $R^2=\frac{8r^2}{9}.$
When $R^2=\frac{8r^2}{9}.$,height of the cone $=r+\sqrt{r^2-\frac{8r^2}{9}}$
$=r+\sqrt{\frac{r^2}{9}}=r+\frac{r}{3}=\frac{4r}{3}.$
Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}.$