A cardboard box of volume 32 cubic inches with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

To find the minimum area of cardboard needed, we need to minimize the surface area of the box. Let x be the length of one side of the square base, and let h be the height of the box. Since the volume of the box is 32 cubic inches, we have: x^2h = 32 We want to minimize the surface area A of the box, given by: A = x^2 + 4xh Using the equation for the volume, we can solve for h in terms of x: h = 32 / x^2 Substituting this into the equation for A, we have: A = x^2 + 4x(32/x^2) = x^2 + 128/x To find the minimum value of A, we take the derivative of A with respect to x and set it equal to zero: dA/dx = 2x - 128/x^2 = 0 Solving for x, we get x = 8, which gives h = 1. Since this value of x makes dA/dx = 0, it is a critical point. We need to check that it's a minimum by checking the second derivative: d^2A/dx^2 = 2 + 256/x^3 At x = 8, this is positive, so we have a minimum at x = 8. Therefore, the minimum area of cardboard needed is: A = x^2 + 4xh = 8^2 + 4(8)(1) = 96 square inches.