The ratio of the height of a cone of maximum | vedclass

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(B) Let the radius of the sphere be $R$ and the diameter be $AE = 2R$.Let the radius of the cone be $x$ and its height be $y$.In the right-angled tria

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$4/3$

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(B) Let the radius of the sphere be $R$ and the diameter be $AE = 2R$.Let the radius of the cone be $x$ and its height be $y$.In the right-angled triangle $BDE$,by the property of chords,$BD^2 = AD 	imes DE$,where $AD = y$ and $DE = 2R - y$.So,$x^2 = y(2R - y)$.The volume of the cone is $V = \frac{1}{3}\pi x^2 y = \frac{1}{3}\pi y(2R - y)y = \frac{1}{3}\pi (2Ry^2 - y^3)$.To find the maximum volume,differentiate $V$ with respect to $y$:$\frac{dV}{dy} = \frac{1}{3}\pi (4Ry - 3y^2)$.Set $\frac{dV}{dy} = 0$:$\frac{1}{3}\pi y(4R - 3y) = 0$.Since $y 
eq 0$,we have $y = \frac{4}{3}R$.Using the second derivative test,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 6y)$.At $y = \frac{4}{3}R$,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 8R) = -\frac{4}{3}\pi R The ratio of the height of the cone to the radius of the sphere is $\frac{y}{R} = \frac{4}{3}$.

The ratio of the height of a cone of maximum volume inscribed in a sphere to its radius is

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