A right circular cone is inscribed in a sphere of radius $3 \sqrt{2}$ units. The radius of the base of the cone so that the volume of the cone is maximum is

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$2 \sqrt{2}$

$2 \sqrt{3}$

answer is C.

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Detailed Solution

Let AB=h

BC=h-R

$\begin{array}{l} r^{2} = 4 h R - h^{2} a n d v = \frac{1}{3} π r^{2} h \\ ∴ v = \frac{1}{3} π (4 h R - h^{2}) h \\ \frac{d v}{d h} = 0 \Rightarrow r = 4 \end{array}$

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