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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 ? - CBSE (Science) Class 12 - Mathematics

#### Question

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

#### Solution

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

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

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

$\text { Squaring both the sides, we get}$

$h^2 + R^2 - 2hR = R^2 - r^2$

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

$\text { Now,}$

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

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

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

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

$\frac{dV}{dh} = 0$

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

$\Rightarrow 4hR = 3 h^2$

$\Rightarrow h = \frac{4R}{3}$

$\text { Now,}$

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

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

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

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

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

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Solution Show that the Height of the Cone of Maximum Volume that Can Be Inscribed in a Sphere of Radius 12 Cm is 16 Cm ? Concept: Graph of Maxima and Minima.
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