Question 1162911
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 72]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ lw\ =\ 308]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{308}{w}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{308}{w}\ +\ w\ =\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 308\ +\ w^2\ -\ 36w\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 36w\ +\ 308\ =\ 0]


Solve the quadratic for *[tex \Large w].  The two roots will be the length and the width.  You decide which is which.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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