Question 624573
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The perimeter of a rectangle is given by *[tex \LARGE P\ =\ 2l\ +\ 2w] and the area of a rectangle is given by *[tex \LARGE A\ =\ lw]


First, we know that the perimeter is equal to 64, so we write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 64]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 32]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 32\ -\ w]


And we know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ lw\ =\ 240]


Make the subsitution from the perimeter equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (32\ -\ w)w\ =\ 240]


A little algebra music, Sammy:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 32w\ +\ 240\ =\ 0]


Solve the quadratic for *[tex \LARGE w], then calculate *[tex \LARGE l\ =\ 32\ -\ w].  Hint:  It factors.


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