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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Solution

$Let the height, radius of base and volume of the cone be h, r and V, respectively . Then, h = R + \sqrt{R^{2} - r^{2}} \Rightarrow h - R = \sqrt{R^{2} - r^{2}} Squaring both the sides, we get h^{2} + R^{2} - 2 h R = R^{2} - r^{2} \Rightarrow r^{2} = 2 h R - h^{2} . . . (1) Now, V = \frac{1}{3} π r^{2} h \Rightarrow V = \frac{π}{3} (2 h^{2} R - h^{3}) [From eq . (1)] \Rightarrow \frac{d V}{d h} = \frac{π}{3} (4 h R - 3 h^{2}) For maximum or minimum values of V, we must have \frac{d V}{d h} = 0 \Rightarrow \frac{π}{3} (4 h R - 3 h^{2}) = 0 \Rightarrow 4 h R = 3 h^{2} \Rightarrow h = \frac{4 R}{3} Now, \frac{d^{2} V}{d h^{2}} = \frac{π}{3} (4 R - 6 h) \Rightarrow \frac{π}{3} (4 R - 8 R) = 0 \Rightarrow \frac{- 4 π R}{3} < 0 So, the volume is maximum when h = \frac{4 R}{3} . \Rightarrow h = \frac{4 \times 12}{3} = 16 c m$

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