Question 978359
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If a rectangle has an area *[tex \Large A] and a perimeter *[tex \Large P], then the length and width of the rectangle are related to the area and perimeter thus:


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


so


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


and then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  l\ =\ \frac{P}{2}\ -\ w]


Then


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


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\ =\ w\left(\frac{P}{2}\ -\ w\right)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\ = -w^2\ +\ \frac{P}{2}w]


And further:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -w^2\ +\ \frac{P}{2}w\ -\ A\ = 0]


Plug in your constants for Area and Perimeter, and then solve the quadratic for *[tex \Large w].  You should get 2 positive roots, one of which will be the width and the other the length.  If you don't get a real number solution, then a rectangle of that area with the given perimeter does not exist.


Hint:  Your quadratic factors, but it will probably take a bit of skull work to find the factors.  You'll probably find your answers more quickly if you use the quadratic formula.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \