. 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 optimize its dimensions. Let's start by visualizing the pyramid and labeling its dimensions.

Let:
- The side length of the square base be x
- The height of the pyramid be h

First, we need to create an equation for the volume of the pyramid in terms of x and h. The volume of a pyramid is given by the formula:

V = (1/3) * base area * height

Since the base of the pyramid is a square, the base area is given by:

Base area = x^2

Substituting this into the volume equation, we have:

V = (1/3) * x^2 * h

Now comes the differentiation part. We need to differentiate V with respect to x (since x is the variable) in order to find the maximum volume. But, before we do that, we need to express h in terms of x.

To do this, let's consider the isosceles triangle faces that are cut away from the square. Each of these triangles has two sides of length x (which are also two sides of the square) and a base side of length b.

Using the Pythagorean theorem, we can find the value of b:

b^2 = x^2 + x^2 [since both sides are of length x]
= 2x^2

Since the base of the triangle is isosceles, it can be split down the middle to form two congruent right triangles. Each of these right triangles will have sides of length b/2, x, and a height of h. Using the Pythagorean theorem, we can find the value of h:

h^2 = (b/2)^2 - x^2
= (2x^2/2)^2 - x^2
= 3x^4/4 - x^2

Now we have an expression for h in terms of x. Let's substitute it back into the volume equation:

V = (1/3) * x^2 * (3x^4/4 - x^2)
V = (x^2 / 4) * (3x^4 - 4x^2)

Simplifying, we obtain:

V = (3x^6 - 4x^4)/4

Now, we can differentiate V with respect to x and set the derivative equal to zero to find the critical points (where the maximum or minimum occurs). Then, we can solve for x:

dV/dx = (18x^5 - 16x^3)/4
= (9x^5 - 8x^3)/2

Setting this equal to zero and solving for x:

(9x^5 - 8x^3)/2 = 0

9x^5 - 8x^3 = 0

Factor out an x^3:

x^3(9x^2 - 8) = 0

This equation is satisfied when x = 0 or 9x^2 - 8 = 0.

Since the side length of the cardboard cannot be zero, we only consider the non-zero solution:

9x^2 - 8 = 0

Solving for x:

9x^2 = 8
x^2 = 8/9
x = √(8/9)

With this value of x, we can find the corresponding value of h by substituting into the expression we derived for h:

h^2 = 3(8/9)^2/4 - (8/9)^2
h^2 = 64/81 - 64/81
h^2 = 0

This implies that the height of the pyramid is zero, which means the volume is also zero.

Therefore, the largest possible volume for the pyramid is zero when the side length of the square cardboard is √(8/9) m.