Posted by **Mary** on Monday, January 16, 2012 at 6:40pm.

consider the collection of all rectangles that have lengths 2in. less than twice their widths. find the possible widths(in inches) of these rectangles if their perimeters are less than 200 in

- pre- calculus -
**Steve**, Tuesday, January 17, 2012 at 12:22pm
let width be x

so length y = 2x-2

p = 2(x+y) = 2x+2(2x-2) = 6x-4

we want p < 200

6x-4 < 200

6x < 204

x < 34

so, as long as the width is less than 34, the length will be less than 66, and the perimeter will be less than 200.

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