To find the maximum area that can be enclosed and separated in this way, we need to use calculus and optimization techniques.
Let's denote the length of the field as L and the width of each divided part as x. Since the field is rectangular, the total length of the fences used would be:
Perimeter of the field = 2L + 2x + 2x = 2L + 4x
From the problem, we know that the total length of the fence available is 500 meters, so we can write the equation:
2L + 4x = 500
Now, we need to express the area of the whole field in terms of L and x. The area (A) of a rectangle is given by:
A = L * x
To maximize the area, we need to find the values of L and x that satisfy the constraint 2L + 4x = 500 and maximize the area A = L * x.
To solve this problem, we can use the method of substitution. Rearrange the constraint equation to solve for L:
2L = 500 - 4x
L = (500 - 4x) / 2
L = 250 - 2x
Now substitute this expression for L into the area equation:
A = (250 - 2x) * x
A = 250x - 2x^2
Next, we differentiate the area equation with respect to x to find the critical points:
dA/dx = 250 - 4x
Set dA/dx equal to zero to find the critical point:
250 - 4x = 0
4x = 250
x = 250/4
x = 62.5
Now we have the value of x, we can substitute it back into the constraint equation to find the value of L:
2L + 4(62.5) = 500
2L + 250 = 500
2L = 250
L = 250/2
L = 125
So, the dimensions of the field that maximize the enclosed and separated area are L = 125 meters and x = 62.5 meters.
Finally, calculate the maximum area using these dimensions:
A = L * x
A = 125 * 62.5
A = 7812.5 square meters
Therefore, the maximum area that can be enclosed and separated in this way with 500 meters of fencing is 7812.5 square meters.