The height of the cone of maximum volume insc | vedclass

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(C) Let the height of the cone be $h$ and the radius of the base of the cone be $r$. Given,the radius of the sphere is $R$. In the right-angled triang

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$\frac{4 R}{3}$

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(C) Let the height of the cone be $h$ and the radius of the base of the cone be $r$. Given,the radius of the sphere is $R$. In the right-angled triangle $	riangle OPB$,where $O$ is the center of the sphere,$P$ is the center of the base of the cone,and $B$ is a point on the circumference of the base of the cone: $R^2 = r^2 + (h - R)^2$ $r^2 = R^2 - (h - R)^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2$. The volume of the cone $V$ is given by: $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2Rh - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$. To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2)$. Setting $\frac{dV}{dh} = 0$: $\frac{\pi}{3} h(4R - 3h) = 0$. Since $h 
eq 0$,we have $h = \frac{4R}{3}$. Checking the second derivative: $\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6h)$. At $h = \frac{4R}{3}$,$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6(\frac{4R}{3})) = \frac{\pi}{3} (4R - 8R) = -\frac{4\pi R}{3} Since the second derivative is negative,the volume is maximum at $h = \frac{4R}{3}$.

The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

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