jmc

We need the same perimeter as a $60$ by $20$ rectangle, but the greatest area we can get. right now the perimeter is $160$. To get the greatest area while keeping a perimeter of $160$, the sides should all be $40$. that means an area of $1600$. Right now, the area is $20\times 60$ which is $1200$. $1600-1200=400$ which is $\boxed{400}$.