Final answer:
To minimize the cost, we need to set up an equation for the cost in terms of the dimensions. We can then differentiate the cost function and set the derivative equal to zero to find the critical points. Solving for the dimensions that minimize the cost, we find that the side length of the top and bottom squares is approximately 3.914 feet and the height of the box is approximately 1.562 feet.
Step-by-step explanation:
To find the dimensions that minimize the cost of the box, we need to set up an equation for the cost in terms of the dimensions. Let's suppose the side length of the top and bottom squares is x and the height of the box is h. The area of each side of the box is 4xh, and the area of the top and bottom is x*x = x^2. The total cost is 2(4xh) + 2(x^2) = 8xh + 2x^2. The volume of the box is x^2 * h = 30. We can solve this system of equations to find the dimensions that minimize the cost.
First, we can write h in terms of x using the volume equation: h = 30/x^2. Substituting this equation into the cost equation, we get C(x) = 8x(30/x^2) + 2x^2 = 240/x + 2x^2.
To minimize the cost, we need to find the critical points of the cost function by taking its derivative. Differentiating with respect to x, we get dC/dx = -240/x^2 + 4x. Setting this derivative equal to zero, we have -240/x^2 + 4x = 0. Multiplying through by x^2, we get -240 + 4x^3 = 0. Solving for x^3, we get x^3 = 240/4 = 60.
Taking the cube root of both sides, we find x = 60^(1/3) ≈ 3.914. Substituting this value back into the equation for h, we find h = 30/(3.914)^2 ≈ 1.562.
Therefore, the approximate dimensions that minimize the cost of the box are x ≈ 3.914 feet for the side length of the top and bottom squares, and h ≈ 1.562 feet for the height of the box.