The maximum volume ( in cubic units) of the cylinder which can be inscribed in a sphere of diameter 6 units is

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$4 \sqrt{3} π$

$h$ $8 \sqrt{3} π$

$12 \sqrt{3} π$

$3 \sqrt{3} π$

answer is A.

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Detailed Solution

Let $h$ be the height, r be the base radius and V be the volume of the cylinder which can be inscribed in a sphere of diameter 6 units

from the diagram $r^{2} + \frac{h^{2}}{4} = 9 \Rightarrow r^{2} = 9 - \frac{h^{2}}{4} .$

$V = π r^{2} h = π h (9 - \frac{h^{2}}{4}) = 9 π h - \frac{π}{4} h^{3} .$

$\frac{d V}{d h} = 9 π - \frac{3 π}{4} h^{2} .$

$∴ \frac{d V}{d h} = 0 \Rightarrow 9 π - \frac{3 π}{4} h^{2} = 0 \Rightarrow h^{2} = 12 \Rightarrow h = 2 \sqrt{3} .$

$\frac{d^{2} V}{d h^{2}} = - \frac{3}{2} π h < 0 \Rightarrow V$ has maximum when $h = 2 \sqrt{3}$

$∴$ Maximum volume $= 9 π (2 \sqrt{3}) - \frac{π}{4} {(2 \sqrt{3})}^{3} = 18 \sqrt{3} π - 6 \sqrt{3} π = 12 \sqrt{3} π .$

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