A box with a square base and open top must have a volume of 62,500 cm3. Find the dimensions of the box that minimize the amount of material used.sides of base     cmheight

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

1. Let's denote the side length of the square base as x (in cm) and the height of the box as h (in cm). The volume V of the box is given by V = x^2 * h. We know that V = 62500 cm^3.

2. From the volume equation, we can express h in terms of x: h = 62500 / x^2.

3. The surface area A of the box (which corresponds to the amount of material used) is given by A = x^2 (for the base) + 4xh (for the sides). Substituting h from step 2, we get A = x^2 + 4x(62500 / x^2) = x^2 + 250000 / x.

4. To minimize A, we need to find its critical points. The derivative of A with respect to x is A' = 2x - 250000 / x^2. Setting A' = 0, we get 2x^3 - 250000 = 0.

5. Solving this equation for x, we get x = (125000)^(1/3) = 50 cm.

6. Substituting x = 50 cm into the equation from step 2, we get h = 62500 / 50^2 = 25 cm.

So, the dimensions that minimize the amount of material used are a square base of side length 50 cm and a height of 25 cm.