. 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 differentiate the volume equation with respect to the variable, in this case, the length of the side of the square base. Let's go step by step:
1. Let's start by considering the square base of the pyramid. The length of each side of the square base is 'a'.
2. Now, let's consider one of the isosceles triangles formed by cutting away a smaller congruent triangle from each corner of the square. The height of this triangle can be labeled as 'h', and the base of the triangle (which is also one side of the square base) can be labeled as 'b'.
3. Based on the properties of an isosceles triangle, we know that the remaining side of the triangle (which is an edge of the pyramid) has the length sqrt(b^2 + h^2).
4. To calculate the volume of the pyramid, we use the formula: V = (1/3) * base area * height.
5. The base area can be found by squaring the length of one side of the square base: base area = a^2.
6. The height of the pyramid can be found by considering the height of the isosceles triangle, which is half of the remaining side length: h = (1/2) * sqrt(b^2 + h^2).
7. Now, we can express the volume in terms of 'a' and 'b': V = (1/3) * (a^2) * ((1/2) * sqrt(b^2 + h^2)).
8. To find the largest volume of the pyramid, we need to find the maximum value of V with respect to 'b'.
9. To differentiate the volume equation with respect to 'b', we apply the power rule and the chain rule of differentiation.
10. Let's differentiate the volume equation: dV/db = (1/3) * (a^2) * (1/2) * (1/2) * (b^2 + h^2)^(-1/2) * 2b.
11. Simplifying the equation: dV/db = (1/3) * (a^2) * b / sqrt(b^2 + h^2).
12. Now, set dV/db equal to zero to find the critical points: (1/3) * (a^2) * b / sqrt(b^2 + h^2) = 0.
13. Simplifying further, we have a = 0 or b = 0.
14. Since a represents the length of the side of the square base, a cannot be zero. Therefore, b must be zero.
15. If b equals zero, the isosceles triangles disappear, and we are left with just a flat square shape. In this case, the volume of the pyramid is also zero.
16. Therefore, the maximum volume the pyramid can have is zero.
In conclusion, the largest volume the pyramid can have, assuming the square piece of cardboard has sides measuring 'a' units, is zero.