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

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+1 vote
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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

BO2 = BP2 + OP2

(12)2 = R2 + (h - 12)2

⇒ 144 = R2 + h2 + 144 - 24h

⇒ R2= 24h - h2

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.

+1 vote
by (35 points)

 Let r and h be the radius and height of cone inscribed in sphere of radius 12cm

In △AOB,122=(h−12)2+r2

∴ r2=24h−h2

volume,

v=31​πr2h

v=31​π(24h−h2)h

∴v=3π​(24h2−h3)

derivating the above equation, we get,

dhdv​=3π​(48h−3h2)

equate dhdv​=0

⇒ 3π​(48h−3h2)=0

solving the above equation, we get,

h=16

Again derivating the above equation, we get,

dh2d2v​=3π​(48−6h)

dh2d2v​→h=16=3π​(48−6(16))<0  

Volume is maximum, when h=16

Hence the height of the cone

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