. 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 the pyramid can have, we need to maximize the volume of the pyramid with respect to the side length of the square piece of cardboard.
Let's start by understanding the dimensions of the pyramid. Since the base of the pyramid is a square, each side of the base has a length of "a".
To find the height of the pyramid, we can consider one of the isosceles triangles on the side face. Let's call the height of the triangle "h". According to the Pythagorean theorem, the length of the base of the triangle (which is equal to half of the side of the square base) is given by:
b = √(a^2 - h^2)
Since the height of the pyramid will be the height of the isosceles triangle, we can express the volume of the pyramid as:
V = (1/3) * (Area of Base) * height
V = (1/3) * (a^2) * h
Now, to find the maximum volume, we need to take the derivative of the volume equation with respect to "h" and set it equal to zero.
dV/dh = (1/3) * (a^2) * (dh/dh)
Simplifying, we get:
dV/dh = (1/3) * (a^2)
Setting this equal to zero:
(1/3) * (a^2) = 0
From this, we can see that the derivative is equal to zero when a^2 = 0. However, since "a" represents the side length of the square, it cannot be zero (as the cardboard wouldn't exist). Therefore, there is no critical value for "h" to find.
To confirm that this is indeed the maximum value, we can look at the second derivative:
d^2V/dh^2 = 0
Since the second derivative is equal to zero, we can conclude that the volume is at a maximum.
Therefore, the largest volume the pyramid can have is when the height "h" is maximized, which occurs when the isosceles triangle on the side face is a right triangle. In this case, the height "h" is equal to "a".
So, the largest volume V would be:
V = (1/3) * (a^2) * a
V = (1/3) * a^3
Thus, the largest volume of the pyramid is equal to (1/3) times the cube of the side length of the square piece of cardboard.