The maximum total area will be square. Let the side of the square be a. The area of the square is therefore a². When the area is divided into two areas by setting the ends of the central dividing fence a distance x from the ends of the nearest corner, the length of the fencing will be 3a+(x+(a-x))+(x+(a-x))=5a=780. So a=156. The area is therefore 156²=24336 sq ft.
To prove that the maximum area is a square let L, the length of the rectangle, be z+y and W, the width of the rectangle be z-y. The area=LW=z²-y². This has a maximum value of z² when y=0. But if y=0, L=W=z, so the rectangle is a square. Also, z-y=W and z+y=L. Add these two equations: 2z=W+L, so z=(L+W)/2 or the average of the length and width of any rectangle. But L+W is half the perimeter P, so z=P/4. So, if we know the length of the perimeter we can work out the length of the side of the square that maximises the enclosed area.