Question 254172
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Let *[tex \Large w] represent the width and let *[tex \Large l] represent the length.


The perimeter is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 30]


Which can be solved for *[tex \Large l]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 15\ -\ w]


(verification of that last step is left as an exercise for the student)


The area is the length times the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ lw\ =\ 50]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (15\ -\ w)w\ =\ 50]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15w\ -\ w^2\ =\ 50]


Rearrange to standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 15w\ +\ 50\ =\ 0]


Just solve the factorable quadratic.  The width will be one of the roots and the length will be the other.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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