Question 732631
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


So


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


Then


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


Substituting to create Area as a function of width:


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


or put another way:


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


Note that this is a quadratic in *[tex \LARGE w] with a negative lead coefficient so the graph is a parabola opening downward.  Hence the vertex represents the value of the independent variable (*[tex \LARGE w]) that yields the maximum area (*[tex \LARGE A(w)]).


The *[tex \LARGE x] coordinate of the vertex of the parabola represented by *[tex \LARGE \phi(x)\ =\ a^2\ +\ bx\ +\ c] is given by *[tex \LARGE -\frac{b}{2a}],  hence the value of the width that yields the maximum area for our rectangle must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ -\frac{\frac{P}{2}}{\,2(-1)}\ =\ \frac{P}{4}].


I leave it as an exercise for the student to calculate the length given that the width is *[tex \LARGE \frac{P}{4}].


J0ohn
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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