What is the height of the cone of maximum value that can be inscribed in a sphere of radius 12cm?

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.

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\[\text { Let the height, radius of base and volume of the cone be h, r and V, respectively . Then, } \]

\[h = R + \sqrt{R^2 - r^2}\]

\[ \Rightarrow h - R = \sqrt{R^2 - r^2}\]

\[\text { Squaring both the sides, we get}\]

\[ h^2 + R^2 - 2hR = R^2 - r^2 \]

\[ \Rightarrow r^2 = 2hR - h^2 ........ \left( 1 \right)\]

\[\text { Now,} \]

\[V = \frac{1}{3}\pi r^2 h\]

\[ \Rightarrow V = \frac{\pi}{3}\left( 2 h^2 R - h^3 \right) ..............\left[ \text {From eq. } \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dV}{dh} = \frac{\pi}{3}\left( 4hR - 3 h^2 \right)\]

\[\text { For maximum or minimum values of V, we must have }\]

\[\frac{dV}{dh} = 0\]

\[ \Rightarrow \frac{\pi}{3}\left( 4hR - 3 h^2 \right) = 0\]

\[ \Rightarrow 4hR = 3 h^2 \]

\[ \Rightarrow h = \frac{4R}{3}\]

\[\text { Now,} \]

\[\frac{d^2 V}{d h^2} = \frac{\pi}{3}\left( 4R - 6h \right)\]

\[ \Rightarrow \frac{\pi}{3}\left( 4R - 8R \right) = 0\]

\[ \Rightarrow \frac{- 4\pi R}{3} < 0\]

\[\text { So, the volume is maximum when h } = \frac{4R}{3} . \]

\[ \Rightarrow h = \frac{4 \times 12}{3} = 16 cm\]

Text Solution

Solution : Let the radius and height of the cone be `r` and `h`. `r^2=24h-h^2` `V=1/3pir^2h` `=1/3pi(24h-h^2 )h` `thereforeV=pi/3(24h^2−h^3)` On differentiating, we get `{dV}/{dh}=pi/3(48h−3h^2)` Put `{dV}/{dh}=0` `impliespi/3(48h−3h^2)=0` `implies h=16` On again differentiating, we get `{d^2V}/{dh^2}=pi/3(48−6h)` `({d^2V}/{dh^2})_{h=16}=pi/3(48−6(16)) <0` ( maxima ) Maximum volume when `h=16cm.` Hence Proved.