The maximum volume of the right circular cone | vedclass

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(B) $\because$ Slant height of the cone,$L = 6$ units. Let the radius be $r$ and height be $h$. Volume $(V) = \frac{1}{3} \pi r^2 h$. Since $L^2 = r^2

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$16 \sqrt{3} \pi$ cu units

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(B) $\because$ Slant height of the cone,$L = 6$ units. Let the radius be $r$ and height be $h$. Volume $(V) = \frac{1}{3} \pi r^2 h$. Since $L^2 = r^2 + h^2$,we have $r^2 = L^2 - h^2 = 36 - h^2$. Substituting this into the volume formula: $V = \frac{1}{3} \pi (36 - h^2) h = \frac{1}{3} \pi (36h - h^3)$. To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (36 - 3h^2)$. Setting $\frac{dV}{dh} = 0$,we get $36 - 3h^2 = 0 \Rightarrow h^2 = 12 \Rightarrow h = 2 \sqrt{3}$ units. Checking the second derivative: $\frac{d^2 V}{dh^2} = \frac{1}{3} \pi (-6h) = -2 \pi h$. At $h = 2 \sqrt{3}$,$\frac{d^2 V}{dh^2} = -4 \sqrt{3} \pi Maximum volume $V = \frac{1}{3} \pi (36 - (2 \sqrt{3})^2) (2 \sqrt{3}) = \frac{1}{3} \pi (36 - 12) (2 \sqrt{3}) = \frac{1}{3} \pi (24) (2 \sqrt{3}) = 16 \sqrt{3} \pi$ cu units.

The maximum volume of the right circular cone with slant height $6$ units is

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