Maximizing enclosed area with fixed fencing along a river

The fencing constraint is $2x + y = 400$, since two sides of length $x$ and one side of length $y$ use all the fencing. Solving for $y$ gives $y = 400 - 2x$. The area to maximize is $A = xy = x(400 - 2x) = 400x - 2x^2$, defined for $0 < x < 200$ so that both $x$ and $y$ stay positive.

Differentiating gives $A'(x) = 400 - 4x$. Setting $A'(x) = 0$ yields $x = 100$, and since $A''(x) = -4 < 0$, this critical point is a maximum. Then $y = 400 - 2(100) = 200$, so the maximum area is $A = 100 \cdot 200 = 20{,}000$ square meters.