. 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

To find the largest volume that the pyramid can have, we need to maximize the volume function with respect to the parameter 'a'.

Step 1: Define the variables and constraints
Let's start by defining the variables:
- 'a' represents the side length of the square piece of cardboard
- 'h' represents the height of the pyramid
- 'V' represents the volume of the pyramid

The constraints in this problem are:
- The base of the pyramid is a square, so the length of each side is 'a'.
- The pyramid is formed from four isosceles triangles, so the height 'h' is the same for all four triangles.

Step 2: Express the volume of the pyramid
The volume of a pyramid can be calculated using the formula:
V = (1/3) * base area * height

The base area of the pyramid is the area of the square base, which is equal to 'a^2'.

Substituting the values, we get:
V = (1/3) * a^2 * h

Step 3: Express the height 'h' in terms of 'a'
To express the height 'h' in terms of 'a', we can use the Pythagorean theorem.

Since the base of the pyramid is square, the height 'h' can be calculated using the formula:
h = sqrt((a^2)/2)

Step 4: Substitute the expression for 'h' in the volume equation
Now, substitute the expression for 'h' obtained in the previous step into the volume equation:
V = (1/3) * a^2 * sqrt((a^2)/2)

Step 5: Differentiate the volume equation
To find the maximum volume, we need to differentiate the volume equation with respect to 'a'.

Differentiating the volume equation, we get:
dV/da = (2/3) * a * sqrt((a^2)/2) + (1/3) * a^2 * (1/2) * (2a/sqrt(2))

Simplifying the expression, we have:
dV/da = (2/3) * a * sqrt((a^2)/2) + (1/3) * a^3 / sqrt(2)

Step 6: Set the derivative equal to zero and solve for 'a'
To find the value of 'a' for which the volume is maximized, set the derivative equal to zero and solve for 'a':
(2/3) * a * sqrt((a^2)/2) + (1/3) * a^3 / sqrt(2) = 0

This is a challenging equation to solve explicitly, so we can use numerical methods or calculators to find the optimal value of 'a'.