The maximum volume i n   c u . m of the right circular cone having slant height 3   m is:

1. $2\sqrt{3} \pi$
2. $3\sqrt{3} \pi$
3. $6 \pi$
4. $\frac{4}{3}\pi$

Solution

Given $l=3$



$\Rightarrow r^2+h^2=l^2=9$

Volume, $V=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi \left(9-h^2\right)h=\frac{1}{3}\pi \left(9h-h^3\right)$

$\frac{dv}{dh}=\frac{1}{3}\pi \left(9-3h^2\right)$

$\frac{dv}{dh}=0\Rightarrow h=\sqrt{3}$

$\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi \left(-6h\right)<0$

$\therefore$ at $h=\sqrt{3}$ , cone has maximum volume

$\therefore$ $V_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{3}\pi \left(9\sqrt{3}-3\sqrt{3}\right)=2\sqrt{3}\pi c{m}^3$

Asked in: JEE Main 2019 (09 Jan Shift 1)

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