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From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
It is less than 5 miles.
width=16 and length=20
No, it is less than 26.2 miles.
It is (W - 6) feet where W is the width measured in feet.