. 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 with respect to the change in one variable, which is the side length of the square base (a).
Let's begin by determining the height of the pyramid. We know that the pyramid has four isosceles triangles as its faces, with two equal side lengths and one different side length. By symmetry, we can take one of the triangles and calculate its height.
Let's consider one of the isosceles triangles. The base of the triangle is the side length of the square (a), and the two equal sides are also of length a. We can bisect the triangle from the apex to the midpoint of the base, creating two congruent right triangles.
Using the Pythagorean theorem, we can find the length of the height (h) of one of the congruent right triangles:
h^2 = a^2 - (a/2)^2
h^2 = a^2 - a^2/4
h^2 = 3a^2/4
h = √(3a^2/4)
h = a√3/2
Now, we can calculate the volume of the pyramid using the formula:
Volume = (1/3) * base area * height
The base area of the pyramid is given by the formula for the area of a square:
Base area = a^2
Substituting the value of the height, we get:
Volume = (1/3) * a^2 * (a√3/2)
Volume = (a^3√3)/6
Now, to find the maximum volume of the pyramid, we can differentiate the volume equation with respect to a and set it equal to zero to find the critical points.
dV/da = (3a^2√3)/6
Setting dV/da = 0, we obtain:
(3a^2√3)/6 = 0
Simplifying, we find:
a^2 = 0
This implies that a = 0, but since we are dealing with lengths, a cannot be zero. Therefore, there are no critical points for the volume equation.
To confirm that this is indeed the maximum volume, we can check the endpoints of the allowed values of a. Since a represents the side length of a square, it cannot be negative, so the only constraint is that a > 0.
At the lower endpoint, when a = 0, the volume is zero. At the upper endpoint, there is no limit set, so we can assume that a is very large.
Taking the limit as a approaches infinity, we find that the volume approaches infinity as well. Therefore, the maximum volume of the pyramid is infinite.
In conclusion, the largest volume the pyramid can have, assuming that the square piece of cardboard has sides measuring a meters, is infinite.