Question 986480
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This problem cannot be solved as written because area is not measured in feet, in other words the statement "the area is 200 ft" is nonsense.  I'm going to go out on a limb and make the assumption that you meant "the area is 200 <i><b>square</b></i> feet".  Don't roll your eyes and say "Whatever"; details like this do make a difference.


The length plus the width of a rectangle is equal to one-half of the perimeter.  Since the perimeter is 100 feet, the sum of the measures of the sides of the rectangle must be 50 feet.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 50\ -\ w\ \ \ \ (1)]


The area is the length times the width, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(l,w)\ =\ lw\ \ \ \ (2)]


But we can substitute from Equation (1) into Equation (2) to get an expression for the area as a function of the width:


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


Since we know that the area is equal to *[tex \Large 200\ ft^2], we can write:


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


Solve for *[tex \Large w] and then calculate *[tex \Large l\ =\ 50\ -\ w]


Hint:  The quadratic does not factor, so you need to use the quadratic formula.


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

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