A farmer wishes to enclose a rectangular plot using 200 m of fencing material. One side of the land borders a river and does not need fencing. What is the largest area that can be enclosed?

1 Answer
Dec 6, 2017

Make the fence have a length parallel to the river of 100m and a width occurring at both ends of the length (perpendicular to the river) of 50m. The area is then 5000m^2

Explanation:

Since one side is against a river only three sides need to be fenced in. So those three sides can be called a Length and two Widths
Thus
L + 2W= 200m, the amount of fencing available.
We wish to maximize the area
A = LW subject to the constraint that
L + 2W= 200
So substituting L = 200 - 2W into the area equation gives:
A = LW = (200 - 2W)W= 200W-2W^2
So
Taking the derivative with respect to W
(dA)/(dW)=200-4W
Setting derivative = 0 to find a max (or min)
200-4W=0 yields 200=4W
And W=200/4=50 and

L = 200-2W which Is L=200-2(50)=100

So the rectangular area should have 2 ends perpendicular to river (the widths by my naming) that are each 50m and a length parallel to river that is100m long.

The resulting area is (50m) (100m)= 5000m^2

Checking the total Length of fenced used; the length is 2w +L
Which is 2(50)+100=200

Checking 2nd derivative
(d^2A)/(d^2W)=-4

Since the 2nd derivative is negative, the curve is concave down and it was a maximum at W=50 as we desired.