Question 278610
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The perimeter of a rectangle is given by:


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



Solving for *[tex \Large l]:


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


The area of a rectangle is given by:


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


Substituting for *[tex \Large l]:


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


Putting the quadratic into standard form:


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


Hence *[tex \Large a\ =\ -1] and *[tex \Large b\ =\ \frac{P}{2}]


and the vertex of the concave down parabola is at


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


Therefore the maximum area is when the width is one-fourth of the perimeter.  If the width is one-fourth of the perimeter, 2 times the width is one-half of the perimeter and then 2 times the length is also one-half of the perimeter (see the perimeter formula).  And then the length must also be one-fourth of the perimeter.  So the maximum area rectangle for ANY given perimeter is a square with a side measure one-fourth of the perimeter.


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