A rectangular box is designed to have a square base and an open top. The volume is to be 2916in.32916⁢in.3What is the minimum surface area that such a box can have?

To solve this problem, we can use calculus and the method of optimization. Here are the steps:

1. Let's denote the side length of the square base as x and the height of the box as h. The volume V of the box is given by V = x^2 * h.

2. We know that the volume is 2916 cubic inches, so we can write the equation as x^2 * h = 2916.

3. We can solve this equation for h to get h = 2916 / x^2.

4. The surface area A of the box (with an open top) is given by A = x^2 + 4xh.

5. Substituting h from step 3 into this equation gives A = x^2 + 4x(2916 / x^2) = x^2 + 11664 / x.

6. To find the minimum surface area, we need to find the derivative of A with respect to x, set it equal to zero, and solve for x. The derivative of A is A' = 2*x - 11664 / x^2.

7. Setting A' equal to zero gives 2x - 11664 / x^2 = 0. Multiplying through by x^2 gives 2x^3 - 11664 = 0.

8. Solving this equation for x gives x = (11664 / 2)^(1/3) = 18.

9. Substituting x = 18 back into the equation for A gives A = 18^2 + 11664 / 18 = 324 + 648 = 972 square inches.

So, the minimum surface area that such a box can have is 972 square inches.