Let O be the centre of a sphere of radius 12 cm and a cone ABC of radius R cm and height h cm is inscribed in the sphere.

AP = AP + OP

⇒ h = 12 + OP

⇒ OP = (h -12)

**Now in right angle ΔOBP,**

**By Pythagoras theorem, we get**

BO^{2} = BP^{2} + OP^{2}

(12)^{2} = R^{2} + (h - 12)^{2}

⇒ 144 = R^{2} + h^{2} + 144 - 24h

⇒ R^{2}= 24h - h^{2}

**Volume of cone,**

**For maximum/minimum value of V, we have**

But height of cone cannot be zero.

Therefore h = 16 cm.

**Now,**

\(\frac{d}{dh}(\frac{dV}{dh})\) \(= \frac{\pi}{3}(48 -6h)\)

⇒ \((\frac{d^2V}{dh^2})_{h=16}\) \(= \frac{\pi}{3}(48 -6\times16)\)

= - 16π < 0.

**Hence, volume of cone is maximum when h = 16 cm.**