Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.

Explanation

Let r be the radius , h the height and S the given surface area of an open cylinder. Then `S = (pi r^(2) + 2pi rh) or h = ((S - pi r^(2))/(2pir))`...(i)
Let V be the volume of the cylinder.
Then, `V = pi r^(2) h = pi r^(2) .((S - pi r^(2))/(2pi r)) = ((rS - pi r^(3))/(2))` [using (i)]
`:. (dV)/(dr) = ((s - 3 pi r^(2))/(2)) and (d^(2)V)/(dr^(2)) = -3pi r`
Now, `(dV)/(dr) = 0 rArr S - 3 pi r^(2) = 0 rArr pi r^(2) + 2pi rh - 3 pi r^(2) = 0 rArr h = r`
And, `((d^(2)V)/(dr^(2)))_(r = h) = -3pi h lt 0`. Thus, r = h is a point of maximum.
So, V is maximum when r = h, i.e., when the height of the cylinder is equal to the radius of its base.