[10 points] A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

  6 inches square by 3 inches high

Step-by-step explanation:

For a given surface area, the volume of an open-top box is maximized when it has the shape of half a cube. If the area were than of the whole cube, it would be 216 in² = 6×36 in².

That is, the bottom is 6 inches square, and the sides are 3 inches high.

Let x and h represent the base edge length and box height, respectively. Then we have ...

  x² +4xh = 108 . . . . box surface area

Solving for height, we get ...

  h = (108 -x²)/(4x) = 27/x -x/4

The volume is the product of base area and height, so is ...

  V = x²h = x²(27/x -x/4) = 27x -x³/4

We want to maximize the volume, so we want to set its derivative to zero.

  dV/dx = 0 = 27 -(3/4)x²

  x² = (4/3)(27) = 36

  x = 6

  h = 108/x² = 3

The box is 6 inches square and 3 inches high.

Comment on maximum volume, minimum area

In the general case of an open-top box, the volume is maximized when the cost of the bottom and the cost of each pair of opposite sides is the same. Here, the "cost" is simply the area, so the area of the bottom is 1/3 the total area, 36 in².

If the box has a closed top, then each pair of opposite sides will have the same cost for a maximum-volume box. If costs are uniform, the box is a cube.