Problems involving fencing rectangular areas all have the same answer: the area is maximized for a given perimeter when the cost of the fence in one direction is equal to the cost of the fence in the orthogonal direction.
The largest area that can be enclosed is 100 ft long (parallel to the river) and 50 ft wide (distance out from the river). It is 5000 ft^2.
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You can let x represent the length of the plot (parallel to the river). Then the area is
.. A = x(200 -x)/2
The is the equation for a parabola with x-intercepts at 200 and 0. The vertex is on the line of symmetry, halfway between those x-intercepts, at x=100. For x=100, the area is
.. A = 100*(200 -100)/2 = 100*50 = 5000 . . . . . ft^2
