Question 696061
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The perimeter of a rectangle is given by *[tex \LARGE P\ =\ 2l\ +\ 2w]


From this we derive *[tex \LARGE l\ =\ \frac{P}{2}\ -\ w]


The area is the length times the width, so area as a function of width is given by:


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


Which can be written


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


Rearranging to standard form and multiplying through by -1:


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


Using the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


where *[tex \LARGE a\ =\ 1], *[tex \LARGE b\ =\ -\frac{P}{2}], and *[tex \LARGE c\ =\ A] we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\frac{P}{2}\ \pm\ \sqrt{\left(-\frac{P}{2}\right)^2\ -\ 4(1)(A)}}{2(1)}]


Just plug in your given values for Perimeter and Area and then do the arithmetic.  One of the roots will be the length (by convention the larger value) and the other will be the width.
 

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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