A fence along a river

Setup. Two sides of length $x$ and one of length $100-2x$, with $0<x<50$. The area is
$$A(x)=x(100-2x)=100x-2x^2.$$

Maximum. $A'(x)=100-4x$ vanishes at $x=25$, with $A'>0$ on $(0,25)$ and $A'<0$ on $(25,50)$. So $A$ is increasing then decreasing on the whole interval, and $x=25$ gives the global maximum.

Answer. $25$ m by $50$ m, area $1250$ m².

Remark. If the field were fenced on all four sides, the best rectangle would be a square. Here the free side lets the field be twice as long as it is wide. In general, with a fence of length $L$, the maximum is at $x=\frac L4$, and the side parallel to the river is $L-2x=\frac L2=2x$: always twice each perpendicular side.