On a piece of paper, draw a square to represent the piece of paper, and another square inside to represent the base of the pyramid. Align the inner square so that the corners are close to the edges of the larger square. The triangular sides of pyramid are formed as in the image.
To figure out size of the base that will maximize the volume of the pyramid, let's call the length of the piece of paper "A," and define "x" as the distance between the edge of the paper and the corner of the base. Then the volume of the pyramid will be a function of x,
V(x) = (1/6)(A-2x)²sqrt(Ax).
(This is obtained using the formula for the volume of a square pyramid, which is (1/3)b²h, where b is the length of the base, and h is the height.)
Next, we maximize the volume equation by taking its derivative with respect to x, setting the derivative equal to zero, and then solving for x. This process leaves us with the critical equation
A² - 12Ax + 20x² = 0,
after a few steps of simplification. When we solve it for x, we get
x = A/10 and x = A/2.
Only the first solution has a practical physical interpretation in the context of the problem, since A/2 is too large to be physically possible. So we get the maximum volume when x is one tenth of A.
As an example, let's construct a pyramid with maximum volume from a 10 by 10 square. Since x = A/10, and A = 10, then we get x = 1. So the corner of the base needs to be 1 unit from edge of the paper. See image for construction diagram. The volume of this pyramid will be about 33.73 cubic units.