A box with an open top has vertical sides, a square bottom, and a volume of 108 cubic meters. if the box has the least possible surface area, find its dimensions. (in your answer leave a space between the number and the unit.)

Let x be the height of the box and y be the length of one side of the base then:-

V = xy^2 = 108

x = 108/y^2

Surface area = y^2 + 4xy

S = y^2 + 4y * 108/y^2

S = y^2 + 432/y

Finding the derivative:-

dS/dy = 2y - 432/y^2 = 0

2y^3 = 432

y^3 = 216

y = 6

Check if this gives a minimum value:-

second derivative = 2 + 864/y^3 which is positive so minimum.

V = xy^2 = 108

36y = 108

y = 3

Answer : - dimensions of the box is 3*6*6 metres