. 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 with respect to the side length of the square, denoted as "a".
Let's first determine the height of the pyramid. Since the pyramid has four isosceles triangular faces, the height will be the same as the slant height of each triangle.
To find the slant height, we can use the Pythagorean theorem. In each isosceles triangle, the two equal sides have length "a", and the base of the triangle is the slant height.
Using the Pythagorean theorem, we have:
(base)^2 + (height)^2 = (slant height)^2
(a/2)^2 + (height)^2 = a^2
Simplifying the equation:
a^2/4 + height^2 = a^2
height^2 = 3a^2/4
height = sqrt(3)a/2
Now, we can calculate the volume of the pyramid. The volume of a pyramid is given by the formula:
Volume = (1/3) * base area * height
Since the base of the pyramid is a square, the base area is given by:
Base area = a^2
Putting it all together, the volume of the pyramid is:
Volume = (1/3) * a^2 * (sqrt(3)a/2)
Simplifying the equation:
Volume = (sqrt(3)/6) * a^3
To find the maximum volume, we need to find the critical points of the volume function with respect to "a".
Differentiating the volume function with respect to "a", we have:
dV/da = (sqrt(3)/2) * a^2
Setting the derivative equal to zero to find the critical point:
(sqrt(3)/2) * a^2 = 0
a^2 = 0
The critical point occurs when "a" equals zero, but since "a" represents the side length of the square cardboard, it cannot be zero. Therefore, there are no critical points.
To determine if the volume is maximum, we need to check the endpoints of the feasible range. The side length "a" should be greater than zero, so there is no lower endpoint.
Therefore, the maximum volume of the pyramid is obtained at the upper endpoint, which is when "a" equals its maximum value. Since a square cannot have a negative side length, the maximum value of "a" is the positive value given in the problem, which is "a = m".
Thus, the largest volume the pyramid can have is:
Volume = (sqrt(3)/6) * (m^3)
Therefore, the maximum volume of the pyramid is (sqrt(3)/6) * (m^3).