A Rectangular Field is to be enclosed by a fence and divided into three parts by another fencer. Find the maximum area that can be enclosed and separated in this way with 500 meters of fencing.
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.