Question

A cone of the greatest volume is inscribed in a sphere. Find the ratio of the radius of the sphere to the altitude of the cone.

• A. $1:1$
• B. $1:\sqrt{2}$
• C. $1:2$
• D. $1:\sqrt{3}$

Answer 1

The correct option is B $1:\sqrt{2}$

$r^2={(h-r)}^2+a^2$

$r^2=h^2+r^2-2hr+a^2$

$a^2=2hr-h^2$

Volume$=\frac{1}{3}\times \pi a^2\times h=\frac{1}{3}\times \pi \times (2hr-h^2)h$

$V=\frac{\pi }{3}(2h^{2}r-h^3)$

$\frac{dV}{dh}=\frac{\pi }{3}(4hr-3h^2)=0$

$4hr=3h^2$

$4r=3h$

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

$a^2=2hr-h^2$

$=2\times \frac{4}{3}{r}^2-\frac{16}{9}{r}^2$

$=\frac{8}{3}{r}^2-\frac{16}{9}{r}^2=\frac{8}{9}{r}^2$

$a=\frac{2\sqrt{2}}{3}r$

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

Ratio$=\frac{a}{h}=\frac{2\sqrt{2}r}{3}\times \frac{3}{4r}$

$\Rightarrow \frac{1}{\sqrt{2}}$