Question

The box is to have a square base and hold 32in^
3. Find the dimensions of the box that can be built with the minimum amount of material. You must justify your answer.

Answer 1

Does it say whether the box is closed or open?

Answer 2

Assume the square base is x inches by x inches and the height is h.
Volume of the box = x * x * h = x^2h = 32 ---- (1)

"minimum amount of material" implies minimum surface area.
Assume the box is open, the surface area A of the box is:
A = x^2 + 4xh ----- (2)

From (1): h = 32 / x^2
From (2): A = x^2 + 4x * 32 / x^2
A = x^2 + 128/x = x^2 + 128x^(-1)
Minimize A by finding the derivative and equating it to zero.
dA/dx = 2x -128x^(-2) = 0
2x - 128/x^2 = 0
multiply by x^2:
2x^3 - 128 = 0
2x^3 = 128
x^3 = 64
x = 4 inches

Put it in (1): 4^2 * h = 32
16h = 32
h = 2

The dimensions of the box are: 4 in. x 4 in. x 2 in.

PS: If it is a closed box the we have to include the surface area of the lid and equation (2) will become A = 2x^2 + 4xh which you can minimize as before.