To find the dimensions of the container with the greatest volume, we can use the method of optimization. Let's denote the length of the base of the container as x and the height as h.
The surface area of the container, excluding the open top, is given by:
2x^2 + 4xh
We know that the total material available is 2000 ft², so we have the equation:
2x^2 + 4xh = 2000
To find the dimensions that maximize the volume, we need to express the volume of the container in terms of a single variable. The volume of a container with a square base is given by:
V = x^2h
We can rewrite the equation for surface area in terms of h:
h = (2000 - 2x^2) / (4x)
Now we can substitute this expression for h in the equation for volume:
V = x^2 * (2000 - 2x^2) / (4x)
Simplifying further:
V = (1/2) * x(1000 - x^2)
To find the maximum volume, we can take the derivative of V with respect to x, set it equal to zero, and solve for x:
dV/dx = (1/2)(1000 - 3x^2)
Setting dV/dx equal to zero and solving for x:
(1/2)(1000 - 3x^2) = 0
1000 - 3x^2 = 0
3x^2 = 1000
x^2 = 1000/3
x = sqrt(1000/3)
Therefore, the length of the base of the container is sqrt(1000/3) and the height can be calculated by substituting this value of x into the equation for h:
h = (2000 - 2(sqrt(1000/3))^2) / (4(sqrt(1000/3)))
Simplifying the expression:
h = (2000 - 2 * (1000/3)) / (4 * sqrt(1000/3))
h = (2000 - 2000/3) / (4 * sqrt(1000/3))
h = (2/3) / (4 * sqrt(1000/3))
h = 2 / (12 * sqrt(1000/3))
h = 1 / (6 * sqrt(1000/3))
Therefore, the dimensions of the container with the greatest volume are approximately:
Length of base = sqrt(1000/3)
Height = 1 / (6 * sqrt(1000/3))