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 the radius and height of the cylinder are 'r' and 'h' respectively.
`:. (2r0^(2) + h^(2) = (2a)^(2)`
`rArr r^(2) = a^(2)- h^(2)/4`…(1)
Volume of cylinder
`V= pi r^(2)h`
` = pi (a^(2) - (h^(2))/4)h` [From eq. (1)]
`=pi(a^(2)h-(h^(3))/4)`
`rArr (dV)/(dh) = pi (a^(2)-(3h^(2))/4)`
For maxima/minima
` (dV)/(dh)=0`
`rArr a^(2) - (3h^(2))/4=0`
`rArr h=(2a)/sqrt3`
Now, `(d^(2)V)/(dh^(2)) = pi (-(3h)/2) lt 0`
Therefore, at `h = (2a)/sqrt3`, the volume of cylinder is msximum.