SOLUTION: The length of a rectangle is 2 cm less than twice the width. Express as an integer the maximum width of the rectangle when the perimeter is less than 66 cm.

Algebra ->  Human-and-algebraic-language -> SOLUTION: The length of a rectangle is 2 cm less than twice the width. Express as an integer the maximum width of the rectangle when the perimeter is less than 66 cm.      Log On


   

Question 1046353: The length of a rectangle is 2 cm less than twice the width. Express as an integer the maximum width of the rectangle when the perimeter is less than 66 cm.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
L = length
W = width


"The length of a rectangle is 2 cm less than twice the width" means that L+=+2W-2


Perimeter of rectangle = 2*(Length+Width)


P+=+2%2A%28L%2BW%29


P+=+2%2A%282W-2%2BW%29 Replace L with 2W-2


P+=+2%2A%283W-2%29 Combine like terms


P+=+6W-4 Distribute


P+=+6W-4


---------------------------------------------------


We are told that the "perimeter is less than 66 cm", so the perimeter P is smaller than 66 meaning that P+%3C+66


P+%3C+66 Start with the inequality


6W-4+%3C+66 Replace P with 6W-4. Solve for W


6W-4%2B4+%3C+66%2B4 Add 4 to both sides


6W+%3C+70 Combine like terms


6W%2F6+%3C+70%2F6 Divide both sides by 6


W+%3C+11.6666666666667 Simplify


Since W is less than 11.6666666666667, this means that the largest W can be is W+=+11


The maximum width is 11 cm