. Folding a Pyramid - A pyramid with a square base and four faces,

each in the shape of an isosceles triangle, is made by cutting away four
triangles from a square piece of cardboard and bending up the resulting
triangles to form the walls of the pyramid. What is the largest volume
the pyramid can have assuming that the square piece of cardboard has
sides measuring a m?

I got everything but the I can't see to find the right solutions, I need help with my differentiation

I must be missing something here. In order for the corners of the triangles to meet, each must have base m/√8 but that gives 4 triangles that just meet in the center if folded up.

For their height to bee enough to form a pyramid, they must overlap, but then the shape is no longer square.

If you cut off 4 small triangles from a square, you get an octagon. How do you fold up "the resulting triangles"?

what formula do you come up with for the volume?

To find the maximum volume of the pyramid, we can use calculus and differentiation. Here's how you can approach this problem:

1. Start by assigning variables to the dimensions of the cardboard and the pyramid. Let the side length of the cardboard square be "a" and the height of the pyramid be "h."

2. Since the pyramid has a square base, the width and length of the base will also be "a" each.

3. The volume of the pyramid can be calculated using the formula: V = (1/3) * base area * height. For a square base, the base area is given by A = a^2.

4. Rewrite the volume equation using the assigned variables: V = (1/3) * a^2 * h.

5. However, we need to relate "a" and "h" as there are constraints given by the problem. When we fold up the triangles to form the pyramid, the height (h) will be equal to half the side length of the square base (a/2).

6. Now we can rewrite the volume equation in terms of a single variable: V = (1/3) * a^2 * (a/2).

7. Simplify the equation: V = (1/6) * a^3.

8. To find the maximum volume, we need to differentiate this equation with respect to "a" and set the derivative equal to zero. Let's do that:

dV/da = (1/6) * 3a^2
= a^2/2

Setting dV/da = 0, we have:
a^2/2 = 0

9. Solve for "a" by taking the square root of both sides:
a = 0

10. Since "a" represents the side length of the square cardboard, we can conclude that the maximum volume of the pyramid is achieved when the side length is zero. However, this doesn't make sense physically.

11. Hence, there is no meaningful solution for the maximum volume in this case. The volume will always be zero for a pyramid formed by folding a square cardboard.

Therefore, the largest possible volume for the pyramid is zero, regardless of the side length of the square cardboard.