Show that a right-circular cylinder of given volume, open at the top, has  minimum total surface area, provided its height is equal to the radius of the base.

Explanation

Let r be the radius and h the height of the cylinder of given volume V.
Then, `V = pi r^(2) h rArr h = (V)/(pir^(2))`.(i)
Let the total surface area be S. Then.
`S = pi r^(2) + 2pi rh = (pi r^(2) + 2pi r .(V)/(pi r^(2)))` [ using (i)]
`:. S = (pi r^(2) + (2V)/(r))`
So, `(dS)/(dr) = (2pi r - (2V)/(r^(2))) and (d^(2)S)/(dr^(2)) = (2pi + (4V)/(r^(3)))`
For a maxima or minima, we have `(dS)/(dr) = 0`
Now, `(dS)/(dr) = 0 rArr 2pi r - (2V)/(r^(2)) = 0 rArr 2pir^(3) = 2V = 2pi r^(2) h rArr r = h`
`((d^(2)S)/(dr^(2)))_(r=h) = (2pi + (4V)/(h^(3))) gt 0`
Thus,` r = h` is a point of minimum.
This shows that the total surface area of the cylinder is minimum when `h = r`, i.e.,
when the height of the cylinder is equal to the radius of the base.