Question

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?

Answer 1

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.