Question 343455
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The area of a rectangle is given by multiplying the length times the width, so in your situation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ lw\ =\ 40]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{40}{w}]


The perimeter is two times the length plus two times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 26]


Subsituting from above:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(\frac{40}{w}\right)\ +\ 2w\ =\ 26]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{80}{w}\ +\ 2w\ =\ 26]


Yuck!  That is uglier than a mud fence.  Not to worry though, we have a bag of lovely cosmetics...


Multiply both sides by *[tex \Large w]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 80\ +\ 2w^2\ =\ 26w]


Add *[tex \Large -26w] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ -\ 26w\ +\ 80\ =\ 0]


Multiply by *[tex \Large \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 13w\ +\ 40\ =\ 0]


Let's see, -5 times -8 is 40 and -5 + -8 is -13.  Can you take it from here?


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