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. The diagonal of the square before the triangles are cut is 24 in. The sides of this square are 12 root 2. The base is square with sides x. (a) Find the height in terms of x. (b) find the max volume and x
To find the height of the pyramid in terms of x, we can use the Pythagorean Theorem. Let's consider one of the isosceles triangles created by cutting the square.
(a) Finding the height:
In this triangle, the base of the triangle is x (which is also a side of the square), and the diagonal of the square, which is also the hypotenuse of the triangle, is 24 inches. Let's denote the height of the triangle as h.
Using the Pythagorean Theorem, we have:
(x/2)^2 + h^2 = 24^2
Simplifying this equation, we get:
x^2/4 + h^2 = 576
Multiplying through by 4 to eliminate the fraction, we have:
x^2 + 4h^2 = 2304
Now, we need to express x in terms of the sides of the square. From the information given, we know that each side of the square is equal to 12√2. Therefore, x = 12√2.
Substituting this value into the equation, we get:
(12√2)^2 + 4h^2 = 2304
288 + 4h^2 = 2304
4h^2 = 2016
Dividing by 4, we get:
h^2 = 504
Taking the square root of both sides, we find:
h = √504
Simplifying the square root, we have:
h = √(4 * 126)
h = √(4 * 9 * 14)
h = 6 * √14
So, the height of the pyramid in terms of x is 6√14.
(b) Finding the maximum volume:
The volume of a pyramid is given by the formula:
Volume = (1/3) * base area * height
In this case, the base is a square with side length x and the height is 6√14.
The base area of the pyramid is given by:
Base area = x * x
So, the volume of the pyramid becomes:
Volume = (1/3) * (x * x) * (6√14)
Now, we need to find the maximum value of the volume. To do this, we can take the derivative of the volume with respect to x, set it equal to zero, and solve for x.
d(Volume)/dx = 0
Simplifying this equation, we have:
(1/3)(2x)(6√14) = 0
12x√14 = 0
x = 0
Since x cannot be zero (as it represents the side length of the square base), this implies that there is no maximum volume.
Therefore, the pyramid does not have a maximum volume.