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(C) Let $H$ be the height and $R$ be the radius of the given cone $V_1$. The volume is $V_1 = \frac{1}{3} \pi R^2 H$.Let the inscribed cone $V_2$ have

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$\frac{4}{27}$

Vedclass · Answer

(C) Let $H$ be the height and $R$ be the radius of the given cone $V_1$. The volume is $V_1 = \frac{1}{3} \pi R^2 H$.Let the inscribed cone $V_2$ have height $h$ and radius $r$. The apex of this cone is at $O$ and its base is at a distance $h$ from $O$. The height of the base of the inscribed cone from the apex $A$ is $H-h$.By similar triangles,$\frac{r}{R} = \frac{H-h}{H} = 1 - \frac{h}{H}$.Thus,$\frac{h}{H} = 1 - \frac{r}{R} \Rightarrow h = H(1 - \frac{r}{R})$.The volume of the inscribed cone is $V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 H(1 - \frac{r}{R}) = \frac{\pi H}{3} (r^2 - \frac{r^3}{R})$.To maximize $V_2$,we differentiate with respect to $r$: $\frac{dV_2}{dr} = \frac{\pi H}{3} (2r - \frac{3r^2}{R}) = 0$.This gives $r(2 - \frac{3r}{R}) = 0$. Since $r 
eq 0$,we have $r = \frac{2R}{3}$.Substituting $r = \frac{2R}{3}$ into the expression for $h$: $h = H(1 - \frac{2R/3}{R}) = H(1 - \frac{2}{3}) = \frac{H}{3}$.Now,$V_2 = \frac{1}{3} \pi (\frac{2R}{3})^2 (\frac{H}{3}) = \frac{1}{3} \pi (\frac{4R^2}{9}) (\frac{H}{3}) = \frac{4}{27} (\frac{1}{3} \pi R^2 H) = \frac{4}{27} V_1$.Therefore,the ratio $\frac{V_2}{V_1} = \frac{4}{27}$.

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