Question 1040508
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*[illustration TimberCutFromLog_(2).jpg]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ 120]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{120}{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 256]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \left(\frac{120}{x}\right)^2\ =\ 256]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ -\ 256x^2\ +\ 14400\ =\ 0]


Let *[tex \Large u\ =\ x^2], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 256u\ +\ 14400\ =\ 0]


Solve the quadratic in *[tex \Large u], then substitute back and take the square root to get *[tex \Large x].  The two positive roots will be your dimensions, since the orientation of the rectangle is arbitrary.


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 \ \ \ \ \ \ \ \ \ \  

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