A picture could look like this:
The perimeter of the fence must be < 200 feet because that's all he has and it makes sense that if he wants to maximize the area, he would want to use all of the fencing available, so P = 200. P also equals 2W + L so 2W + L = 200
Since area is defined as L•W and we want to maximize that area, then we will want to consider the graph of the area as a function of either length or width and find the largest area available considering the constraint of a 200 foot perimeter.
So, since 2W + L = 200, we can say L = 200 - 2W and we can plug "200 - 2W" in for "L" in our A = L•W equation.
A = L•W
A = (200 - 2W)W
A = -2W2 + 200W
This is a quadratic equation that graphs as an inverted parabola (with a maximum at the vertex). So we want to find that maximum. The vertex of a parabola is found using -b/2a so WVERTEX = -200/[(2)(-2)] = 50
If W = 50, then we can plug that into L = 200 - 2W to find L
L = 200 - 2(50) = 100
So the dimensions are 50 x 100 and that yields 5000 sq ft.
Just for grins, let's pick a width of 40 feet just to see what the area would be. L = 200 - 2(40) = 120 so the area would be 40•120 = 4800 sq ft. This supports our solution that the max is 5000 sq ft.