SOLUTION: The length of a rectangle is five times its width. If the perimeter is at most 96 centimeters, what is the greatest possible value for the width?

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Question 798486: The length of a rectangle is five times its width. If the perimeter is at most 96 centimeters, what is the greatest possible value for the width?
Found 2 solutions by waynest, MathTherapy:
Answer by waynest(281) About Me  (Show Source):
You can put this solution on YOUR website!
length = 5w
width = w
p = 96
p = 2w + 2L
2w + 2(w + 5) = 96
2w + 2w + 10 = 96
4w + 10 - 10 = 96 - 10
4w = 86
4w/4 = 86/4
w = 21.5
check:
2(21.5) + 2(21.5 + 5( = 96
43 + 2(26.5) = 96
43 + 53 = 96
96 = 96

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The length of a rectangle is five times its width. If the perimeter is at most 96 centimeters, what is the greatest possible value for the width?

Let width be W

Then length = 5W

Therefore, 2%28W+%2B+5W%29+%3C=+96

2%28W+%2B+5W%29+%3C=+2%2848%29

W+%2B+5W+%3C=+48

6W+%3C=+48

Width, or W+%3C=+48%2F6, or W+%3C=+8

Greatest possible width: highlight_green%288%29 cm

You can do the check!!

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