Question 1120369
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If the perimeter is 64, then *[tex \Large 2l\ +\ 2w\ =\ 64] which is to say *[tex \Large l\ =\ 32\ -\ w]


Since the area is given by the length times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (32\ -\ w)w\ =\ 247]


Solve the quadratic for *[tex \Large w] and then calculate *[tex \Large 32\ -\ w].  Hint:  Discard the negative root of the quadratic since it is highly unlikely that the desired rectangle has a negative number dimension.
								
								
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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