Question 249886
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Start with a width of 1.  If the width is 1, 2 times 1 is 2, 18 minus 2 is 16, half of 16 is 8, so: 1 by 8. Area equals 1 times 8 is 8


2 times 2 is 4, 18 minus 4 is 14, 14 divided by 2 is 7, so: 2 by 7, area is 2 times 7 is 14


and so on...


When you get to the point where the width is larger than the length, you can stop because you have exhausted all the possibilities.  The greatest area will be the rectangle that is closer in shape to a square than any of the others -- that means that the difference between the length and the width is the smallest possible amount.  That is because for any given perimeter, the rectangle dimensions that give the greatest area is a square with sides measuring the perimeter divided by 4.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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