Question 221882
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The perimeter of a rectangle is found by:


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


so the length of a rectangle in terms of perimeter and width is:


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


The area of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ lw]


Substituting we can create an Area function in terms of width for any given perimeter:


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


Putting this quadratic function in standard form results in:


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


This graphs to a parabola opening downward, hence the vertex is a maximum.  The vertex of a parabola represented by *[tex \Large y = ax^2 + bx + c] has an *[tex \Large x]-coordinate of *[tex \Large \frac{-b}{2a}], hence the vertex for our Area function is:


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


Hence, the width giving the maximum area is *[tex \Large \frac{P}{4}].  But twice the width is then *[tex \Large \frac{P}{2}], hence for any given perimeter, twice the length must be *[tex \Large \frac{P}{2}] as well.  Therefore, for the maximum area, the length must be equal to the width and the maximum area shape is a square with each side measuring one-fourth of the perimeter.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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