Question

With exactly 2700 square inches of cardboard, we wish to construct a box width 2x, depth x, height 2x. We would like to maximize the volume, V, the box can hold. Which values of width, depth, and height fulfill our objective?

Answer 1

To maximize the volume of the box, we need to optimize the dimensions of the box such that we use all of the available cardboard. Let's call the width of the box 2x, the depth of the box x, and the height of the box 2x.

The surface area of the box can be calculated as:

SA = 2(2x * x) + 2(2x * 2x) + 2(x * 2x)
SA = 4x^2 + 8x^2 + 4x^2
SA = 16x^2

We know that the cardboard we have is 2700 square inches, so:

16x^2 = 2700

Solving for x:

x^2 = 168.75

x ≈ 12.99 inches

Now that we have the value of x, we can calculate the dimensions of the box:

Width = 2x = 2(12.99) ≈ 25.98 inches
Depth = x = 12.99 inches
Height = 2x = 2(12.99) ≈ 25.98 inches

So the box with dimensions 25.98 x 12.99 x 25.98 inches will have a volume that maximizes the use of the given cardboard. We can calculate the volume as:

V = width * depth * height
V = 25.98 * 12.99 * 25.98
V ≈ 8,171.66 cubic inches