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**

- All categories
- Pre-Algebra Answers 12,297
- Algebra 1 Answers 25,208
- Algebra 2 Answers 10,418
- Geometry Answers 5,148
- Trigonometry Answers 2,634
- Calculus Answers 5,990
- Statistics Answers 3,013
- Word Problem Answers 10,026
- Other Math Topics 6,546

81,280 questions

85,416 answers

2,167 comments

68,898 users

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.