Answer:
To maximize the area enclosed by a rectangular fence with a fixed perimeter and one side adjucent to a barn, we can use calculus or a symmetry argument. Since one side of the enclosure does not need fencing (because it's adjucent to the barn), we can consider the fence as a three-sided rectangle.
Let's denote the length of the rectangle as L and the width as W. The perimeter P of the three-sided enclosure is given by:
P = L + 2W
Since the farmer has 200 feet of fencing, we have:
200 = L + 2W
We can solve for L:
L = 200 - 2W
The area A of the rectangle is given by:
A = L * W
Substitute L from the perimeter equation into the area equation:
A = (200 - 2W) * W
A = 200W - 2W^2
To find the maximum area, we take the derivative of A with respect to W and set it to zero:
dA/dW = 200 - 4W = 0
Solving for W gives:
4W = 200
W = 50 feet
Now, we can find L using the perimeter equation:
L = 200 - 2W
L = 200 - 2(50)
L = 200 - 100
L = 100 feet
So, the dimensions that will give the greatest area are 100 feet for the length (parallel to the barn) and 50 feet for the width (perpendicular to the barn). This results in a maximum area of:
A = L * W
A = 100 feet * 50 feet
A = 5000 square feet
This is the maximum area that can be enclosed with 200 feet of fencing when one side is adjucent to a barn.