. 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, side length a) and set it equal to zero to find the critical points. Let's go step by step.
Step 1: Determine the volume equation:
The volume V of a pyramid with a square base of side length a and height h can be calculated using the formula V = (1/3) * base area * height.
Let's determine the base area first:
The base of the pyramid is a square, so its area is a * a = a^2.
Step 2: Substitute the base area and height into the volume equation:
V = (1/3) * a^2 * h
Step 3: Express the height in terms of a:
Since the pyramid has isosceles triangular faces, the height h will be half the length of the slant height s. We can use the Pythagorean theorem to express h in terms of a:
a^2 + (s/2)^2 = s^2
Simplifying the equation:
a^2 + s^2/4 = s^2
4a^2 + s^2 = 4s^2
4a^2 = 3s^2
s^2 = (4/3) * a^2
s = (2/√3) * a
Since the slant height s is equal to the height h/2, we can write:
h = (2/√3) * a / 2
h = (√3/3) * a
Step 4: Substitute the height back into the volume equation:
V = (1/3) * a^2 * (√3/3) * a
V = (√3/9) * a^3
Step 5: Differentiate the volume equation with respect to a:
dV/da = (√3/9) * 3a^2
dV/da = (√3/3) * a^2
Step 6: Set the derivative equal to zero and solve for a:
(√3/3) * a^2 = 0
a = 0
We have found a critical point at a = 0, but this is not relevant to our problem since it represents a degenerate case (a collapsed or non-existent pyramid). We need to find the maximum volume, so we are looking for a positive value of a.
Hence, the largest volume the pyramid can have is at a critical point where a > 0.
Please note that this solution assumes that all four isosceles triangles have the same dimensions.
To find the largest volume the pyramid can have, we need to maximize the volume function with respect to the side length of the square cardboard.
First, let's start by visualizing the pyramid. We have a pyramid with a square base and four triangular faces. The base of the pyramid is a square with side length a, and each face is an isosceles triangle with two sides of length a and a height of h. To find the height, we can use the Pythagorean theorem:
h = √(a^2 - (a/2)^2)
h = √(a^2 - a^2/4)
h = √(3a^2/4)
h = √(3/4)a
The volume of a pyramid is given by the formula:
V = (1/3) * base area * height
V = (1/3) * a^2 * √(3/4)a
V = (1/3) * √(3*a^4)/4
Now, let's differentiate the volume function with respect to a to find its critical points:
dV/da = 0
To differentiate the function, we can use the power rule and the chain rule:
dV/da = (1/3) * (1/2) * (3*a^4/4)^(-1/2) * (12*a^3/4)
dV/da = (1/4) * (3*a^4/4)^(-1/2) * (12*a^3/4)
dV/da = 3 * a^3 * (3*a^4/4)^(-1/2)
Simplifying further:
dV/da = 3 * a^3 * (16/9a^4)^(-1/2)
dV/da = 3 * a^3 * (16/9)^(-1/2) * a^(-2)
dV/da = 3 * a^3 * (16/9)^(-1/2) * a^(-2)
dV/da = (3/2) * (16/9)^(-1/2) * a
To find the critical points, we solve for a when dV/da = 0:
(3/2) * (16/9)^(-1/2) * a = 0
Since the constant term (3/2) * (16/9)^(-1/2) is non-zero, we can conclude that there are no critical points in the interval (0, inf).
Therefore, to maximize the volume, we need to consider the endpoints of the interval. The largest volume will occur either at a = 0 or as a approaches infinity.
However, since we are dealing with a real-world scenario and a represents the side length of the square cardboard, a cannot be 0. Therefore, we consider the limit as a approaches infinity.
lim V as a approaches infinity = lim ((1/3) * √(3*a^4)/4) as a approaches infinity
lim V as a approaches infinity = ∞
Thus, the largest possible volume for the pyramid is infinity.
Therefore, the largest volume the pyramid can have is unbounded.