Question

Prove that the volume of largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.

Answer 1

Let R = radius of sphere

r = base radius cone

R+R = height cone

V = value cone

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

By Pythagorean theorem

$r^2=R^2-R^2$

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

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

$\frac{dV}{dR}=0$ for crical point.

$\frac{dV}{dR}=\frac{1}{3}\pi (R^2-2Rh-3h^2)=0$

$R^2-2Rh-3h^2=0$

$(R-3h)(R+h)=0$

$R=\frac{R}{3},-R$

h must be +ve

$h=\frac{R}{3}$

$\frac{d^{2}V}{d{R}^2}$ for Nature of point

$\frac{d^{2}V}{d{R}^2}=\frac{\pi }{3}(-2R-6h)<0$

So, if will be max.

$r^2=R^2-h^2=R^2-\frac{R^2}{3}=\frac{8}{3}{R}^2$

$V_{max}=\frac{\pi }{3}\left(\frac{8}{3}{R}^2\right)(R+\frac{R}{3})$

$=\frac{8}{27}\pi R^3\left(\frac{4}{3}\right)$

$=\frac{32}{81}\pi R^3$

$V=\left(\frac{8}{27}\right)\left(\frac{4}{3}\pi R^3\right)$

Hence proved.