Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

Let a cone of radius r cm and height h cm is inscribed in a sphere of radius 12 cm.

In right triangle AOB,

(12)² =  (h – 12)² + r²

r² =24h – h²

Now, V = (1/3)πr²h

V =  (1/3)π(24h – h²)h  (Substituting the value of r²)

V = (1/3) π(24h² – h³)

dV/dh = (1/3) π (48h – 3h²)

For maximum volume: dV/dh = 0

Therefore, we get

(1/3) π (48h – 3h²) = 0

48h – 3h² = 0

h = 16.

Also, d²V/dh² = 1/3 π (48 – 6h)

(d²V/dh² )_{h = 16} =  1/3 π (48 – 96) < 0.

Therefore, for h = 16, volume is maximum.

Hence, height of the cone  of maximum volume, which can be inscribed in a sphere of radius 12 cm is 16 cm.