Question 608165
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The perimeter of a rectangle is given by:


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


Which can be rearranged to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{P\ -\ 2w}{2}]


The area, given by length times width, can then be represented as a function of the width for a given perimeter thus:


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


This function is a quadratic that can be put into standard form thus:


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


The graph of such a quadratic is a parabola that is concave down (the lead coefficient is less than zero), meaning that the vertex represents a maximum.  The *[tex \Large w] coordinate of the vertex of this parabola is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-b}{2a}\ =\ \frac{-\frac{P}{2}}{2(-1)}\ = \frac{P}{4}]


Therefore the width that gives the greatest area rectangle for any given perimeter is the perimeter divided by 4.  The two widths are therefore half of the perimeter, hence the two lengths must also be half of the perimeter, and therefore the shape is a square.


Divide your given perimeter by 4 and then square the result.


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