show that the cone of greatest volume which can be inscribed in a given sphere has an altitude =2/3RD of - Maths -

Question

show that the cone of greatest volume which can be inscribed in
a given sphere has an altitude =2/3RD of diameter of sphere

Santosh Kumar · Accepted Answer

Let be the centre of the sphere of radius r.

Also let ACB be the height of the cone

ACB = h units

BC = h – r

Let s be the radius of the cone

Now, volume of the cone, $v = \frac{1}{3} π s^{2} h ... (1)$

In ABCD,

$\Rightarrow r^{2} = (h - r)^{2} + s^{2} \Rightarrow s^{2} = 2 h r - h^{2}$

Putting value of s² in equation (1)

$v = \frac{1}{3} π (2 h r - h^{2}) = \frac{π}{3} (2 h^{2} r - h^{3}) \frac{d v}{d h} = \frac{π}{3} (4 h r - 3 h^{2})$

For maximum volume,

$\frac{d v}{d h} = 0 \Rightarrow h = \frac{4}{3} r \Rightarrow h = \frac{4}{3} (\frac{d}{2}) = \frac{2 d}{3}$

where, d = diameter of the sphere

Hence, cone of greatest volume which can be inscribed in a given sphere has an altitude $= \frac{2}{3}$ (diameter) of the sphere.

show that the cone of greatest volume which can be inscribed in a given sphere has an altitude =2/3RD of diameter of sphere.