Question 516683
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The rectangle with the greatest area for a given perimeter is a square with sides that measure the perimeter divided by 4.


The perimeter of a rectangle is given by:


*[tex \LARGE \text{          }\math P = 2l + 2w \ \ \Rightarrow\ \ 2l = P - 2w \ \ \Rightarrow\ \ l = \frac{P}{2} - w]


The area of a rectangle is given by:


*[tex \LARGE \text{          }\math A = lw]


Substituting from the perimeter equation:


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


This function graphs to a parabola opening downward meaning that the vertex is a maximum.


<u>Algebra solution:</u>  The vertex is the maximum, so for a parabola with equation *[tex \LARGE f(x)\ =\ ax^2\ +\ bx\ +\ c], the vertex is at *[tex \LARGE \left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)].  For this parabola:  *[tex \LARGE \left(\frac{-\frac{P}{2}}{2(-1)},f\left(\frac{-\frac{P}{2}}{2(-1)}\right)\right)\ =\ \left(\frac{P}{4},f\left(\frac{P}{4}\right)\right)]


<u>Calculus solution:</u> The maximum value of the function, hence the maximum area, is where the value of the first derivative is equal to zero:


*[tex \LARGE \text{          }\math \frac{d}{dw}A(w)\ =\  -2w + \frac{P}{2}]


Set equal to zero:


*[tex \LARGE \text{          }\math -2w + \frac{P}{2} = 0 \ \ \Rightarrow\ \ 2w = \frac{P}{2} \ \ \Rightarrow\ \ w = \frac{P}{4}]


Hence, the maximum area rectangle for a given perimeter is a square with sides of length one-fourth of the perimeter.


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