jmc

Let the sides perpendicular to the barn be of length $x$. Notice that there are a total of $1200/5=240$ feet of fencing. Therefore, the side parallel to the barn has length $240-2x$, so the area to be maximized is $240x-2x^2$. Completing the square results in $-2(x-60{)}^2+7200$, which is maximized when $x=60$. Therefore, the side parallel to the barn has length $240-2(60)=\boxed{120}$ feet.