Let's assume that the dimensions of the square base of the box are x by x, and the height of the box is h.
Since the box is a rectangular solid with a square base, the volume of the box is given by the formula V = length x width x height. In this case, we have a square base, so the length and width are the same and equal to x. Therefore, the volume equation becomes V = x^2 * h.
Given that the volume of the box is 12 ft^3, we can write the equation as:
x^2 * h = 12
Next, we need to find the surface area of the box to determine the cost. The surface area (SA) can be calculated by adding the areas of the six sides of the box. Since the square base has four sides, each with an area of x * h, and the other two sides have areas of x^2, the total surface area is given by:
SA = 4(x * h) + 2(x^2)
Based on the problem, the cost of the top of the box, which is one of the four sides with an area of x * h, is twice as expensive as the other sides, which are the other three sides and the bottom with areas of x * h. Let's assume the cost of the latter is C, then the cost of the top is 2C.
The cost equation can be written as:
Cost = C * (4(x * h) + 2(x^2)) + 2C * (x * h)
Since we want to find the most economical dimensions, we need to minimize the cost function. We can solve for x and h by substituting the volume equation (x^2 * h = 12) into the cost equation.
Substituting x^2 * h = 12 into the cost equation, we get:
Cost = C * (4(x * h) + 2(x^2)) + 2C * (x * h)
Cost = C * (4(x * h) + 2(x^2)) + 2C * (12 / h)
Now, let's take the derivative of the cost function with respect to h and set it to zero to find a critical point. Differentiating the cost function, we get:
dCost / dh = C * (4x - 24 / h^2) - 24C / h^2
Setting dCost / dh = 0, we can solve for h:
0 = C * (4x - 24 / h^2) - 24C / h^2
Simplifying the equation, we get:
4x - 24 / h^2 = 24 / h^2
4x = 48 / h^2
x = 12 / h^2
Now, substitute the value of x into the volume equation (x^2 * h = 12):
(12 / h^2)^2 * h = 12
144 / h^4 * h = 12
144 / h^3 = 12
144 = 12h^3
h^3 = 144 / 12
h^3 = 12
h = 2
Substituting h = 2 into x = 12 / h^2, we get:
x = 12 / (2^2)
x = 12 / 4
x = 3
Therefore, the most economical dimensions of the box are a square base with dimensions 3 ft by 3 ft and a height of 2 ft.