I assume that you have made a sketch. That is the first step to solving the problem.
Imagine making the type of enclosure described. Let the sides parallel to the fence in the middle have length $x$, and let the other two sides of the rectangle each have length $y$. Write $x$ beside the two sides and the fence in the middle that have length $x$, and write $y$ beside the two sides that have length $y$.
From the picture, we see that the amount of fencing used is $3x+2y$. if we want to make the enclosure large, itt is clear that we are best off using all the available fencing. Thus we must have $3x+2y=900$. We want to maximize the area $xy$, given that $3x+2y=900$.
Now proceed as has been suggested for similar problems. We have $y=\frac{1}{2}(900-3x)$.
So we want to maximize $f(x)=\frac{1}{2}(x)(900-3x)$.
If we want to use calculus, note that $f(x)=450x -\frac{3}{2}x^2$. So $f'(x)=450-3x$. The derivative is $0$ at $x=150$.
The maximum area is therefore $f(150)$, which is not hard to compute. I would prefer to find $y$ from $3x+2y=900$. If $x=150$ then from $3x+2y=900$ we get $y=225$. Thus the maximum area that can be enclosed is $(150)(225)$.
Remark: If we want to be very fussy (and sometimes it can be important to be), we can note that $0\le x\le 300$ (where $x=0$ and $x=300$ do not give "real" fields). Our function $f$ attains a maximum in the interval $0\le x\le 300$. But obviously $f(0)=f(300)=0$, so the maximum is not attained at an endpoint. Thus the maximum is reached at a place where $f'(x)=0$. There is only one such place, namely $x=150$, so the maximum must be reached there. Thus our calculation does yield the maximum area.
The second point first... your objection is perfectly sensible, and if Paul only wanted to consider rectangles, he should have said so in the question. In fact, forming the fence into a circle - or to be more precise, a semicircle - will give a larger area. (A square, on the other hand, is a special kind of rectangle and therefore is not excluded by Paul's method of solution, though it doesn't happen to be the right answer in this case.)
The restrictions on $y$ arise from considering the physical meaning of the problem. You can't (in this question) have a length less than zero for the vertical side, so the minimum possibility is $y=0$. You can't have a length less than zero for the horizontal side either, and a little thought will show you that this means $y$ can't be greater than $250$.
I think perhaps you are not recognising that "no sides to the fence" is different from "sides of length zero".
Best Answer
Let $r$ be the length of the rectangle along the river and $s$ be the length of the other side. Thus, we have to:
Max $rs$
Subject to:
$r + 2s = 1400$
Substitute for $r$ back into the objective function and we get the problem reformulated as:
Max $(1400-2s) s$
You can find the value of $s$ either via calculus or by completing the square.