Final answer:
Using calculus optimization techniques, the maximum volume of the box is shown to be c^3/6√3 cubic units by expressing the height in terms of the base side value and given surface area, differentiating the volume with respect to the base side, and solving for the critical point.
Step-by-step explanation:
To prove that the maximum volume of an open box with a square base and made from a given quantity of cardboard with area c^2 square units is c^3/6√3 cubic units, we can use optimization techniques from calculus. Let the side of the square base of the box be x and the height of the box be h. The surface area of the four sides of the box plus the base is x^2 + 4xh = c^2. We want to maximize the volume of the box, which is V = x^2h.
We can express h in terms of x and c^2 using the surface area equation, h = (c^2 - x^2) / (4x). Substituting this into the volume expression, we get V = x^2 * (c^2 - x^2) / (4x), which simplifies to V = 1/4 * (c^2x - x^3).
To maximize this volume, we take the derivative with respect to x, set it equal to zero, and solve for x.
The critical point that maximizes the volume will be when x = c/√6. Substituting this back into the volume equation, we get the maximum volume V = c^3/6√3.