2(L+W)=64 5/6"=389/6".

L+W=389/12. L=389/12-W, so the relationship can be plotted on a straight line graph.

389/12-W>0 so W<389/12", or W,L<32 5/12".

If W=L the rectangle is a square so L=W=16 5/24". 16 5/24≤{L W}<32 5/12.

Area is variable=LW=L(389/12-L)=389L/12-L^2 which is maximum when L=W=389/24 (area=262.71 sq in approx.)

2(length+width)=64 5/6

Length+width=389/12

Length = 389/12-width

So,

width and length are grater than 389/12

Finding **Aera of Rectangle**

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Insufficient information. There are an infinite number of rectangles with this perimeter which is equal to twice the sum of length and width. Also need area or relative lengths of sides.