Answer:
- 2(L +W) ≤ 600
- W ≤ 200
- L ≥ 2W
Explanation:
We assume the problem wording means the length is to be at least 2 times as long as the width. (Longer than usually refers to a difference, not a scale factor.)
If we let "W" and "L" represent the width and length, respectively, then we can translate the problem statement to ...
2(L + W) ≤ 600 . . . . . . the perimeter is twice the sum of length and width
W ≤ 200 . . . . . . . . . . . . the width is at most 200 inches
L ≥ 2W . . . . . . . . . . . . . the length is at least twice the width