Question 807497
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For a given perimeter, a square is the greatest area rectangle that can be constructed.  *[tex \LARGE 4(x\ -\ 1)\ =\ 2x\ +\ 2(x\ -\ 2)] so your square and rectangle have the same perimeter.


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


so


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


Since area is length times width


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


Since this is a quadratic function with a negative lead coefficient, the graph is a downward-opening parabola, hence the function value at the vertex represents a maximum value of the function.


The independent variable (*[tex \Large w]) coordinate of the vertex is:


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


Therefore the maximum area is obtained when the width of the rectangle is the perimeter divided by 4.  If the perimeter divided by 4 is the width, 2 times the width is the perimeter divided by 2, and then 2 times the length must also be the perimeter divided by 2.  The four sides are equal in measure, and the maximum area figure is a square.


John
*[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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