If a right circular cone having maximum volum | vedclass

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(A) Let $R = 3 \, cm$ be the radius of the sphere.Let $h$ be the height and $b$ be the base radius of the inscribed cone.From the geometry of the sphe

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$8\sqrt{3} \pi$

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(A) Let $R = 3 \, cm$ be the radius of the sphere.Let $h$ be the height and $b$ be the base radius of the inscribed cone.From the geometry of the sphere,the relationship between $h, b,$ and $R$ is given by $(h-R)^2 + b^2 = R^2$.Thus,$b^2 = R^2 - (h-R)^2 = 2hR - h^2$.The volume of the cone is $V = \frac{1}{3} \pi b^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.To maximize volume,we find $\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2) = 0$.This gives $h(4R - 3h) = 0$. Since $h 
eq 0$,we have $h = \frac{4R}{3} = \frac{4(3)}{3} = 4 \, cm$.Now,$b^2 = 2(4)(3) - (4)^2 = 24 - 16 = 8$,so $b = \sqrt{8} = 2\sqrt{2} \, cm$.The slant height $l = \sqrt{h^2 + b^2} = \sqrt{4^2 + (2\sqrt{2})^2} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6} \, cm$.The curved surface area is $A = \pi b l = \pi (2\sqrt{2}) (2\sqrt{6}) = 4\pi \sqrt{12} = 4\pi (2\sqrt{3}) = 8\sqrt{3} \pi \, cm^2$.

If a right circular cone having maximum volume is inscribed in a sphere of radius $3 \, cm$,then the curved surface area (in $cm^2$) of this cone is

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