I'll tell you more: when you fix the perimeter of the rectangle, the one with the maximum area is always the square with that perimeter. Here's the proof.
Given a perimeter 2P, all rectangles with that perimeter have sides x and y such that
[tex] 2(x+y)=2P \iff x+y = P \iff x = P-y [/tex]
The area is the product of the dimensions, so
[tex] A = xy = (P-y)y = -y^2+Py [/tex]
The maximum of this parabola is found by setting its derivative to zero:
[tex] -2y + P = 0 \iff y = \dfrac{P}{2} [/tex]
which implies
[tex] x = P-y = P-\dfrac{P}{2} = \dfrac{P}{2} [/tex]
So, the maximum area is achieved when x=y, i.e. when the rectangle is actually a square.
So, the square with perimeter is 3131 has side length [tex] \frac{3131}{4} = 782.75 [/tex]
