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.
Let the two inside parallel walls each have length $x$. Let the sides of the rectangle perpendicular to these each have length $y$.
Then the total area enclosed is $xy$. The amount of fencing used is $4x+2y$. This is to be $900$, since it is clear that it is best to use up all the fencing.
So we want to maximize $xy$, under the constraint $4x+2y=900$.
Thus $y=450-2x$, and we want to maximize $x(450-2x)$.
Because of the physical situation, we need $x\ge 0$ and $y\ge 0$. This means $x\le 225$.
So mathematically, we want to minimize $f(x)=450x-2x^2$, where $0\le x\le 225$.
This can be done by standard tools, such as calculus or completing the square.
Best Answer
Not correct. You forgot to take into account the extra fence that goes across the land.
So, you need to start with $2x +3y=120$