Here r = 3 cos$$\theta $$

and $${h \over 2}$$ = 3 sin$$\theta $$

$$ \Rightarrow $$ h = 6 sin$$\theta $$

We know, Volume of cylinder

V = $$\pi $$ r² h

$$ \Rightarrow $$ V = $$\pi $$ (9cos²$$\theta $$)(6 sin$$\theta $$)

$$ \Rightarrow $$ V = 54$$\pi $$ (sin$$\theta $$ - sin³$$\theta $$)

For maxima/minima of volume,

$${{dV} \over {dh}} = 0$$

$$ \Rightarrow $$ cos$$\theta $$ - 3sin²$$\theta $$ cos$$\theta $$ = 0

$$ \Rightarrow $$ cos$$\theta $$(1 - 3sin²$$\theta $$) = 0

$$ \Rightarrow $$ cos$$\theta $$ = 0

$$ \Rightarrow $$ $$\theta $$ = $${\pi \over 2}$$ (not possible)

or 1 - 3sin²$$\theta $$ = 0

$$ \Rightarrow $$ sin$$\theta $$ = $${1 \over {\sqrt 3 }}$$

$$ \therefore $$ h = 6 sin$$\theta $$ = $${6 \over {\sqrt 3 }}$$ = $$2\sqrt 3 $$