The slant height of a right circular cone is | vedclass

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(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$. Given the slant height $\ell = 3 	ext{ cm}$,we have the relation $\ell^2 = r^2 + h^

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$\sqrt{3} 	ext{ cm}$

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(D) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$. Given the slant height $\ell = 3 	ext{ cm}$,we have the relation $\ell^2 = r^2 + h^2$,which implies $r^2 = 9 - h^2$. Substituting $r^2$ into the volume formula: $V = \frac{1}{3} \pi (9 - h^2) h = 3 \pi h - \frac{\pi}{3} h^3$. To find the maximum volume,we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = 3 \pi - \pi h^2$. Setting $\frac{dV}{dh} = 0$,we get $3 \pi = \pi h^2$,which gives $h^2 = 3$,so $h = \sqrt{3} 	ext{ cm}$ (since height must be positive). Checking the second derivative: $\frac{d^2V}{dh^2} = -2 \pi h$. At $h = \sqrt{3}$,$\frac{d^2V}{dh^2} = -2 \sqrt{3} \pi Since the second derivative is negative,the volume is maximum when $h = \sqrt{3} 	ext{ cm}$.

The slant height of a right circular cone is $3 \text{ cm}$. The height of the cone for maximum volume is

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