Question 1189873
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The perimeter of a rectangle is given by *[tex \Large P\ =\ 2l\ +\ 2w]


So the length for a fixed perimeter as a function of the width is *[tex \Large l\ =\ \frac{P}{2}\ -\ w]


Then since the area is length times width, the area as a function of the width with a fixed perimeter is *[tex \Large A(w)\ =\ \frac{P}{2}w\ -\ w^2].


Given a perimeter of 400 yards and a width of *[tex \Large x], the function you seek is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x)\ =\ 200x\ -\ x^2]


Given that an area less than zero is absurd, the domain of *[tex \Large A] is the set of all *[tex \Large x] such that *[tex \Large A(x)] is non-negative, namely the closed interval between the two zeros of the function. I leave it as an exercise for the student to find the endpoints of the domain interval.

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