. 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 of the pyramid, we need to maximize the volume function with respect to the variable "a" (the side length of the square cardboard).

Let's start by setting up the problem and defining our variables:
- Side length of the square cardboard: a
- Base area of the pyramid (square base): A = a^2
- Height of the pyramid: h (unknown)
- Volume of the pyramid: V = (1/3) * A * h

Now, our goal is to find the maximum value of V. To do this, we can use calculus, specifically differentiation.

Step 1: Rewrite the volume equation in terms of a single variable.
You can substitute A (the base area) in terms of a:
V = (1/3) * A * h
= (1/3) * (a^2) * h
= (1/3) * a^2 * h

Step 2: Determine the variable to differentiate with respect to.
In this case, we want to maximize the volume, so we need to differentiate with respect to the variable "a."

Step 3: Differentiation
Differentiate the volume function V with respect to a using the product rule:
dV/da = (1/3) * (2a) * h [Differentiation of a^2 with respect to a]

Step 4: Set the derivative equal to zero and solve for "a."
To find the maximum volume, we set dV/da = 0 and solve for "a":
(1/3) * (2a) * h = 0
2ah = 0
Since h cannot be zero (that would result in a flat pyramid), we can conclude that a = 0.

Step 5: Determine if it is a maximum or a minimum.
To determine whether we have found a maximum or minimum, we need to consider the second derivative of the volume function.

Take the derivative of the derivative (dV^2/da^2) with respect to "a":
d^2V/da^2 = 2h

Since the second derivative is positive (2h > 0), we have a minimum value at the point where a = 0. This result indicates that there is no maximum volume for the pyramid.

In other words, no matter the side length of the square cardboard, we cannot create a pyramid with a maximum volume.