. 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 involved, which is the height of the pyramid. Let's denote the height of the pyramid as h.
The volume V of a pyramid with a square base is given by the formula: V = (1/3) * base area * height.
In this case, the base area is the area of the square, which is given by: A = a^2, where a is the side length of the square cardboard.
We are given that the side length of the cardboard is given as a, so the base area is a^2.
Now, we can express the volume as a function of height: V(h) = (1/3) * a^2 * h.
To maximize the volume, we need to find the critical points of the volume function. In calculus, we can do this by taking the derivative of the volume function with respect to h and setting it to zero. Then, solve for h to find the critical points.
Differentiating the volume function with respect to h:
dV(h)/dh = (1/3) * a^2 * (dh/dh) = (1/3) * a^2.
Let's set this derivative equal to zero and solve for h:
(1/3) * a^2 = 0.
Multiplying both sides by 3:
a^2 = 0.
Since a^2 can never equal zero, there are no critical points.
However, in this case, we need to find the maximum volume within a given constraint, which is the side length of the cardboard, a. Since the cardboard has a finite length, the height of the pyramid cannot exceed this length. Therefore, the maximum volume occurs when the height of the pyramid is equal to the side length of the cardboard.
So, the largest volume the pyramid can have is V_max = (1/3) * a^2 * a.
Simplifying this expression:
V_max = (1/3) * a^3.
Therefore, the largest volume the pyramid can have is (1/3) * a^3 cubic units.
Note: This result assumes that the cardboard is folded perfectly with no overlap or gaps.