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} + r^{2}

r^{2} =24h – h^{2}

Now, V = (1/3)πr^{2}h

V = (1/3)π(24h – h^{2})h (Substituting the value of r^{2})

V = (1/3) π(24h^{2} – h^{3})

dV/dh = (1/3) π (48h – 3h^{2})

For maximum volume: dV/dh = 0

Therefore, we get

(1/3) π (48h – 3h^{2}) = 0

48h – 3h^{2} = 0

h = 16.

Also, d^{2}V/dh^{2} = 1/3 π (48 – 6h)

(d^{2}V/dh^{2} )_{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.

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