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# Solution for Show that the Cone of the Greatest Volume Which Can Be Inscribed in a Given Spher Has an Altitude Equal to 2 3 of the Diameter of the Sphere. - CBSE (Commerce) Class 12 - Mathematics

#### Question

Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $\frac{2}{3}$ of the diameter of the sphere.

#### Solution

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

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

$\Rightarrow \left( h - R \right)^2 = R^2 - r^2$

$\Rightarrow h^2 + R^2 - 2hr = R^2 - r^2$

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

$\text { Volume of cone } = \frac{1}{3}\pi r^2 h$

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

$\Rightarrow V = \frac{1}{3}\pi\left( 2 h^2 R - h^3 \right)$

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

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

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

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

$\Rightarrow 4hR = 3 h^2$

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

$\text { Substituting the value of y in eq } . \left( 1 \right), \text { we get }$

$x^2 = 4\left( r^2 - \left( \frac{r}{\sqrt{2}} \right)^2 \right)$

$\Rightarrow x^2 = 4\left( r^2 - \frac{r^2}{2} \right)$

$\Rightarrow x^2 = 4\left( \frac{r^2}{2} \right)$

$\Rightarrow x^2 = 2 r^2$

$\Rightarrow x = r\sqrt{2}$

$\text { Now,}$

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

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

$\Rightarrow \frac{d^2 V}{d h^2} = \frac{- 4\pi R}{3} < 0$

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

$\Rightarrow h = \frac{2}{3}\left( \text { Diameter of sphere } \right)$

$\text { Hence proved }.$

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Solution for question: Show that the Cone of the Greatest Volume Which Can Be Inscribed in a Given Spher Has an Altitude Equal to 2 3 of the Diameter of the Sphere. concept: Graph of Maxima and Minima. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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