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

The radius and h is the height of cylinder then from figure
`r^(2)+(h^(2)//4)=a^(2)` or `r^(2)=a^(2)-(h^(2))/(4)`
Now `V=pir^(2)h=pi(a^(2)-(1)/(4)h^(2))h=pi(a^(2)h-(1))/(4)^(3)`
`(dv)/(dh)=pi(a^(2)-(3))/(4)h^(2)=0` for maxmium or minimu
This gives h =`(2//sqrt(3))` a for which `d^(2)v//dh^(2)=-6h//4lt0`
Hence V is maximum when `h=2a//sqrt(3)`