Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius `R` is `(2R)/(sqrt(3))` .

Explanation

Let 'r' be the radius and 'h' be the height of the cylinder.
`:. (2r)^(2)+h^(2)=(2R)^(2)`

` rArr r^(2) = R^(2) - h^(2)/4` ….(1)
Volume of cylinder
` V = pi r^(2)h = pi (R^(2)-(h^(2))/4)h`
` = pi(R^(2)h-h^(3)/4)`
` rArr (dV)/(dh) = pi (R^(2) - (3h^(2))/4)` [From eqn. (1)]
For maxima/minima` (dV)/(dh) = 0`
` rArr R^(2) - (3h^(2))/4 = 0 rArr h= (2R)/sqrt3`
Now `(d^(2)V)/(dh^(2)) = pi (-(3h)/2) lt 0 `
`:. at h = (2R)/sqrt3` the volume of cylinder is maximum.