SOLUTION: The perimeter of a rectangular sign is not to exceed 50 feet. The length is to be twice the width. What widths will meet these conditions? The only formula I know is: P=2(l+w)

Question 620398: The perimeter of a rectangular sign is not to exceed 50 feet. The length is to be twice the width. What widths will meet these conditions?
The only formula I know is: P=2(l+w) I dont understand how this formula solves this problem.....please help me. Your time and efforts will be greatly appreciated. Thank You.
Answer by kingme18(98) (Show Source): You can put this solution on YOUR website!
Since the length is to be twice the width, you can say l = 2w (length equals two times width). That means that l is the same as 2w...so in your formula, instead of l, you can say 2w: P = 2(2w + w)
Since perimeter is 50 ft, plug that in for P and solve:
50=2(2w+w)
50=2(3w)
50=6w
8.333...=w
w is 8.333... and l is twice that, or 16.666...
Since this is in terms of feet, we can write 8.333... ft as 8 ft, 4 inches (because .333... is 1/3, and 1/3 of a foot is 4 inches) and 16.666... ft as 16 ft, 8 inches.
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