Let x be the radius of base, h be the height and S be the surface area of the closed right circular cylinder whose volume is V which is given to be constant.
Then πr2h = V
\(\therefore h = \frac{V}{\pi r^2} = \frac{A}{x^2} ....(1)\)
where A = V/π, which is constant.
\(\therefore\) by the second derivative test, S is minimum when
Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.
