Drum tight containers is designing an open-top, square-based, rectangular box that will have a volume of 62.5 in cubed. what dimensions will minimize surface area? what is the minimum surface area?

Let the square base of the container be of side s inches and the height of the container be h inches, then
Surface are of the container, A = s^2 + 4sh
For minimum surface area, dA / ds + dA / dh = 0
i.e. 2s + 4h + 4s = 0
6s + 4h = 0
s = -2/3 h

But, volume of container = 62.5 in cubed
i.e. s^2 x h = 62.5
(-2/3 h)^2 x h = 62.5
4/9 h^2 x h = 62.5
4/9 h^3 = 62.5
h^3 = 62.5 x 9/4 = 140.625
h = cube root of (140.625) = 5.2 inches
s = 2/3 h = 3.47

Therefore, the dimensions of the square base of the container is 3.47 inches and the height is 5.2 inches.

The minimum surface area = s^2 + 4sh = (3.47)^2 + 4(3.47)(5.2) = 12.02 + 72.11 = 84.13 square inches.