A box with a square base and no top is to be built with a volume of 4000 in3. Find the dimensions of the box that requires the least amount of material. How much material is required at the minimum?

V = x^2 h = 4000 so h = 4000/x^2

A = x^2 + 4 x h
A = x^2 + 4 x (4000/x^2)
A = x^2 + 16,000/x
dA/dx = 0 at max or min
0 = 2 x -16000/x^2

x^3 = 8000

x = 20
now do the rest

To find the dimensions of the box that requires the least amount of material, we need to minimize the surface area of the box.

Let's assume the length of one side of the square base is "x" inches, and the height of the box is "h" inches.

The volume of the box is given as 4000 in³. Since the base is a square, the base area is x² in². The volume of the box can be expressed as the product of the base area and the height: x² * h = 4000.

To minimize the surface area, we need to find an equation that only depends on one variable. So we'll express h in terms of x: h = 4000/x².

The surface area of the box (excluding the top) is the sum of the area of the base and the four sides. The area of the base is x² in², and the area of each side is x * h, which is x * (4000/x²) = 4000/x in².

The total surface area is: surface area = x² + 4 * (4000/x) = x² + 16,000/x in².

To find the dimensions of the box that require the least amount of material, we need to find the critical points of the surface area equation. We'll take the derivative of the equation and set it equal to zero:

d(surface area)/dx = 2x - 16,000/x².

Setting this equal to zero and solving for x:

2x - 16,000/x² = 0
2x = 16,000/x²
2x³ = 16,000
x³ = 8,000
x = 20.

So, the length of one side of the square base is 20 inches.

To find the height, we can substitute x into the volume equation: x² * h = 4000
20² * h = 4000
400 * h = 4000
h = 10.

Therefore, the dimensions of the box that require the least amount of material are 20 inches by 20 inches by 10 inches.

To find the amount of material required at the minimum, we need to calculate the surface area using the dimensions we found:

Surface area = x² + 4 * (4000/x) = 20² + 4 * (4000/20) = 400 + 4 * 200 = 1200 in².

So, at the minimum, 1200 square inches of material are required.