Let sqrt{3} be the radius and frac{pi}{3} be | vedclass

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(D) Let the radius of the cone be $R = \sqrt{3}$ and the semi-vertical angle be $\alpha = \frac{\pi}{3}$.The height of the cone is $H = R \cot(\alpha)

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$3$

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(D) Let the radius of the cone be $R = \sqrt{3}$ and the semi-vertical angle be $\alpha = \frac{\pi}{3}$.The height of the cone is $H = R \cot(\alpha) = \sqrt{3} \cot(\frac{\pi}{3}) = \sqrt{3} 	imes \frac{1}{\sqrt{3}} = 1$.Let the radius of the inscribed cylinder be $r$ and its height be $h$.By similar triangles,$\frac{R-r}{h} = \frac{R}{H} = \frac{\sqrt{3}}{1} = \sqrt{3}$.Thus,$h = \frac{R-r}{\sqrt{3}} = \frac{\sqrt{3}-r}{\sqrt{3}} = 1 - \frac{r}{\sqrt{3}}$.The volume of the cylinder is $V = \pi r^2 h = \pi r^2 (1 - \frac{r}{\sqrt{3}}) = \pi (r^2 - \frac{r^3}{\sqrt{3}})$.To maximize $V$,we find $\frac{dV}{dr} = \pi (2r - \frac{3r^2}{\sqrt{3}}) = \pi (2r - \sqrt{3}r^2) = 0$.Since $r 
eq 0$,we have $2 - \sqrt{3}r = 0$,so $r = \frac{2}{\sqrt{3}}$.The height of the cylinder is $h = 1 - \frac{1}{\sqrt{3}} (\frac{2}{\sqrt{3}}) = 1 - \frac{2}{3} = \frac{1}{3}$.

Let $\sqrt{3}$ be the radius and $\frac{\pi}{3}$ be the semi-vertical angle of a given cone. Then the height of the right circular cylinder of maximum volume that can be inscribed in the given cone is

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