. 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
Sure! To find the largest volume the pyramid can have, we need to maximize the volume function with respect to the variable involved. In this case, the variable is the side length of the square, denoted as "a."
Let's start by finding the relationship between the side length of the square base (a) and the height of the pyramid (h). Since the triangular faces are isosceles, the height of each triangular face will be equal. Let's call this height "h".
To determine the height of the triangular face, we can use the Pythagorean theorem. Let's consider one of the four triangles. The hypotenuse of each triangle is the side length of the square base (a), and the base, which is half of the side length of the square, can be written as (a/2). Therefore, using the Pythagorean theorem, we can express the height (h) as follows:
h^2 = a^2 - (a/2)^2
h^2 = a^2 - (a^2)/4
h^2 = (3/4) * a^2
h = √((3/4) * a^2)
h = (√3/2) * a
Now, we have an expression for the height (h) in terms of the side length of the square base (a).
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 will be equal to the square of the side length (a^2). So, the volume (V) of the pyramid is:
V = (1/3) * a^2 * h
V = (1/3) * a^2 * [(√3/2) * a]
V = (√3/6) * a^3
To find the largest volume, we need to find the critical points of the volume function, which correspond to its maximum or minimum values. So, let's find the derivative of the volume function with respect to "a" and set it equal to zero:
dV/da = (√3/6) * 3a^2
dV/da = (√3/2) * a^2 = 0
Setting (√3/2) * a^2 = 0 gives us a critical point at a = 0.
To find whether this critical point is a maximum, minimum, or neither, we need to examine the second derivative of the volume function. Let's take the derivative again:
d^2V/da^2 = (√3/2) * 2a
For a critical point to be a maximum, the second derivative must be negative. Plugging in a = 0 into the second derivative, we find:
d^2V/da^2 = (√3/2) * 2 * 0 = 0
Since the second derivative is equal to zero, we cannot determine the nature of the critical point using the second derivative test.
However, since the domain of the problem is restricted to positive values of "a", we can see that the volume of the pyramid will increase as "a" increases. Hence, the maximum volume of the pyramid will be obtained when a is as large as possible within the restrictions of the problem.
Therefore, the largest volume the pyramid can have is given by:
V = (√3/6) * (a^3)
where "a" represents the side length of the square base.