<PRE><B>Question 402284</B>
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Let&#39;s solve this one in general, that is for any given perimeter.


Let &lt;b&gt;&lt;i&gt;P&lt;/i&gt;&lt;/b&gt; represent the fixed perimeter.


Let &lt;b&gt;&lt;i&gt;w&lt;/i&gt;&lt;/b&gt; represent the width of the rectangle.


Let &lt;b&gt;&lt;i&gt;l&lt;/i&gt;&lt;/b&gt; represent the length of the rectangle.


The perimeter of a rectangle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P = 2l + 2w]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l = \frac{P - 2w}{2}]


The area of a rectangle is the length times the width so a function for the area in terms of the width is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w) = \left(\frac{P - 2w}{2}\right)w]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w) = \frac{Pw - 2w^2}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w) = \frac{P}{2}w - w^2]


&lt;b&gt;&lt;u&gt;Algebra Solution:&lt;/u&gt;&lt;/b&gt;


The area function is a parabola, opening downward, with vertex at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-\frac{P}{2}}{-2} = \frac{P}{4}]


Since the parabola opens downward, the vertex represents a maximum value of the area function.  The value of the width that gives this maximum value is one-fourth of the given perimeter.  Therefore, the shape must be a square, and the area is the width squared.


&lt;b&gt;&lt;u&gt;Calculus Solution:&lt;/u&gt;&lt;/b&gt;


The area function is continuous and twice differentiable over its domain, therefore there will be a local extrema wherever the first derivative is equal to zero and that extreme point will be a maximum if the second derivative is negative at that point.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w) = \frac{P}{2}w - w^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{dw} = \frac{P}{2} - 2w]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{P}{2} - 2w = 0 \ \ \Rightarrow\ \ w = \frac{P}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{dw^2} = -2]


Therefore the maximum area is obtained when


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w = \frac{P}{4}]


And that maximum area is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\left(\frac{P}{4}\right) = \frac{P}{2}\left(\frac{P}{4}\right) - \left(\frac{P}{4}\right)^2 = \left(\frac{P}{4}\right)^2 = w^2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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