Question 985619
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The perimeter of a rectangle is twice the length plus twice the width.  Therefore, the sum of the length and the width is half the perimeter.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 50]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 50\ -\ w]


Since the area of a rectangle is the length times the width, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ 50w\ -\ w^2]


If the area must be 200, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -w^2\ +\ 50w\ -\ 200\ =\ 0]


Solve the quadratic for *[tex \Large w], then calculate *[tex \Large 50\ -\ w].


Use the quadratic formula because the quadratic equation does not factor.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \