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.

Question 1114438: 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 Theo(13342) (Show Source): You can put this solution on YOUR website!
L = length of the rectangle.
W = width of the rectangle.
P = perimeter of the rectangle.

P = 2L + 2W, which is equal to 2 * (L + W).

you are given that L = 2W - 2.

therefore P = 2 * (L + W) becomes P = 2 * (2W - 2 + W) which becomes P = 2 * (3W - 2) which becomes P = 6W - 4

you are given that P < 66, therefore 6W - 4 < 66.

add 4 to both sides of this inequality to get 6W < 70.

divide both sides of this inequality by 6 to get W < 70/6.

70/6 = 11.6666666...

if you rounded up, you would say the maximum width was 12 cm.

however, a width of 12 cm leads to a length of 2 * 12 - 2 = 22 cm and the perimeter would then be 2 * (12 + 22) = 2 * 34 = 68, which is greater than 66.

rounding down would get you a maximum width of 11.

with a width of 11, the length would be 2 * 11 - 2 = 20 and the perimeter would be 2 * (20 + 11) = 2 * 31 = 62 which is less than 66.

i would say round down to 11 to get to the maximum integer width.

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