The maximum volume (in m^3) of the right circ | vedclass

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(D) Let the slant height be $l = 3 \, m$,radius be $r$,and height be $h$.We know that $l^2 = r^2 + h^2$,so $r^2 = l^2 - h^2 = 9 - h^2$.The volume of t

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$2\sqrt{3}\pi$

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(D) Let the slant height be $l = 3 \, m$,radius be $r$,and height be $h$.We know that $l^2 = r^2 + h^2$,so $r^2 = l^2 - h^2 = 9 - h^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.Substituting $r^2$,we get $V(h) = \frac{1}{3} \pi (9 - h^2) h = \frac{1}{3} \pi (9h - h^3)$.To find the maximum volume,differentiate $V$ with respect to $h$:$\frac{dV}{dh} = \frac{1}{3} \pi (9 - 3h^2)$.Setting $\frac{dV}{dh} = 0$,we get $9 - 3h^2 = 0$,which implies $h^2 = 3$,so $h = \sqrt{3}$.Now,calculate the maximum volume:$V = \frac{1}{3} \pi (9 - 3) \sqrt{3} = \frac{1}{3} \pi (6) \sqrt{3} = 2\sqrt{3}\pi \, m^3$.

The maximum volume (in $m^3$) of the right circular cone having slant height $3 \, m$ is

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