Question 1023080
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To the nearest 5 years? What are you talking about?


Let a rectangle have a perimeter *[tex \Large P], then if the length is *[tex \Large l] and the width is *[tex \Large w], we can say that *[tex \Large l\ +\ w\ =\ \frac{P}{2}], and conclude that *[tex \Large l\ =\ \frac{P}{2}\ -\ w]


Since we know that the area of the rectangle is length times width, we can write an expression for area as a function of width by saying:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ -w^2\ +\ \frac{P}{2}w]


Since the vertex of the parabola *[tex \Large f(x)\ =\ ax^2\ +\ bx\ +\ c] is at the point *[tex \Large \left(x_v,y_v\right)] where:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v = \frac{-b}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v = f(x_v) = f\left(\frac{-b}{2a}\right)]


So, for the area function we derived above, the maximum value of the function is where:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w_{max} = \frac{-\frac{P}{2}}{-2}\ =\ \frac{P}{4}]


Which is to say, you get the maximum area when the width is exactly one-fourth of the perimeter.  The only way for the width to be exactly one-fourth of the perimeter is for the rectangle to have four equal measure sides, i.e. a square.


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

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

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