Question 273321
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The first thing we know is that *[tex \Large l\ =\ 2w]


The next thing we know is that the perimeter of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


Substituting we discover that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 6w\ \Rightarrow\ w\ =\ \frac{P}{6}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 3l \Rightarrow\ l\ =\ \frac{P}{3}]


The area of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ lw]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(P)\ =\ \left(\frac{P}{3}\right)\left(\frac{P}{6}\right)\ =\ \frac{P^2}{18}]


Check the answer for reasonability:


Consider a 2' by 4' rectangle.


The area is 8 ft².  The perimeter is 2 times 4 plus 2 times 2 equals 8 plus 4 equals 12'


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(12)\ =\ \frac{12^2}{18}\ =\ \frac{144}{18}\ =\ 8]


Seems to work.



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