A rectangular box is to have a square base and volume of 78 feet cubed. If the material for the base costs $.30 per square feet, the material for the sides costs $.10 per square foot, and the material for the top costs $.20 per square foot. Determine the dimensions of the box that can be constructed at minimum cost.

If the dimensions are x,z, with z being the height, then

x^2 z = 78, so z=78/x^2

The cost c is

c(x,z) = .30x^2 + 4(.10xz) + .20x^2
But, z=78/x^2, so
c(x) = .30x^2 + .4x(78/x^2) + .20x^2
c(x) = .50x^2 + 31.2/x

Now that's not easy to do algebraically, but with a little calculus we have

dc/dx = x - 31.2/x^2 = (x^3-31.2)/x^2
dc/dx=0 when x^3-31.2=0, or
x=3.148
z=78/3.148^2 = 7.870

So, the box has is 3.148x3.148x7.870

To determine the dimensions of the box that can be constructed at minimum cost, we need to find the dimensions that minimize the cost function. Let's denote the side length of the square base as x and the height of the box as h.

Given that the volume of the box is 78 ft^3, we have the equation: x^2 * h = 78.

We need to minimize the cost function, which is given by: C = A_base * cost_base + A_sides * cost_sides + A_top * cost_top, where A_base is the area of the base, A_sides is the combined area of the four sides, and A_top is the area of the top.

The area of the base is A_base = x^2.
The combined area of the four sides is A_sides = 4 * x * h.
The area of the top is A_top = x^2.

Given the costs per square foot, we have: cost_base = $0.30/ft^2, cost_sides = $0.10/ft^2, and cost_top = $0.20/ft^2.

Substituting the values into the cost function, we have:
C = x^2 * ($0.30/ft^2) + 4 * x * h * ($0.10/ft^2) + x^2 * ($0.20/ft^2).

Now we can express the cost function solely in terms of x and h:
C = 0.3x^2 + 0.4xh + 0.2x^2.

To minimize the cost, we can take the partial derivatives of C with respect to x and h, and set them equal to zero:

∂C/∂x = 0.6x + 0.4h = 0.
∂C/∂h = 0.4x = 0.

From the second equation, we can see that x = 0, which is not valid. Therefore, we can divide the first equation by 0.2 and simplify:

3x + 2h = 0.

Now we can solve this system of equations. Solve the first equation for x in terms of h:

x = -2h/3.

Substitute this expression for x in the equation x^2 * h = 78:

(-2h/3)^2 * h = 78,
4h^3/9 = 78,
4h^3 = 702,
h^3 = 175.5,
h ≈ 6.32 ft.

Substitute the value of h back into x = -2h/3:

x = -2(6.32)/3,
x ≈ -4.21 ft.

However, we cannot have negative dimensions, so we take the absolute value:

x ≈ 4.21 ft.

Therefore, the dimensions of the box that can be constructed at minimum cost are approximately 4.21 ft by 4.21 ft by 6.32 ft.