1 Find the maximum volume of a square based frustum truncate

1) Find the maximum volume of a square based frustum (truncated pyramid).

2) Given a square sheet, find distance d and angle a which will maximize the volume of the frustum (truncated pyramid).

Solution

Volume of a truncated square pyramid is given by:

V = h (a² + ab + b²) /3

First, we use the fact that the volume of a pyramid is (1/3)B²H, where B is the length of the base and H is the height of the pyramid. Using a few applications of the pythagorean theorem, we can see that

B = L/sqrt(2) - sqrt(2)X
H² = [sqrt(L²/4 + x²)]² - (L/2 - X)²
H = sqrt(LX)

So the volume is given by the equation

V = (1/3)(sqrt(LX))(L/sqrt(2) - sqrt(2)X)²

By taking the derivative of V and solving dV/dX = 0, we will get the optimal value of X:

V = [0.5L^5/2X^1/2 - 2L^3/2X^3/2 + 2L^1/2X^5/2]/3
dV/dX = [0.25L^5/2X^-1/2 - 3L^3/2X^1/2 + 5L^1/2X^3/2]/3

[0.25L^5/2X^-1/2 - 3L^3/2X^1/2 + 5L^1/2X^3/2]/3 = 0
0.25L² - 3LX + 5X² = 0
X = [3L ± sqrt(9L² - 5L²)/10
X = L/10 and L/2

The solution X = L/2 corresponds to a pyramid with no volume, therefore, the unique solution is L/10. This means that in order to construct a square pyramid with the largest possible volume, you should make X equal to one tenth of the side length.
answer.