Question 1167080
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Area of a rectangle:  *[tex \Large A\ =\ lw]


Perimeter of a rectangle:  *[tex \Large P\ =\ 2l\ +\ 2w]


Solve the perimeter function for *[tex \Large l]


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


So the area as a function of the width given a fixed perimeter is:


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


The graph of the Area function is a concave down parabola because of the negative coefficient on the w squared term.


The maximum area is obtained at the vertex of the parabola.  The value of the independent variable (w here) at the vertex is the additive inverse of the coefficient on the first degree term divided by 2 times the coefficient on the squared term:


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


So the maximum area occurs when the width is equal to one-fourth of the perimeter, in other words when the length and the width are equal so that the rectangle is a square.


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


I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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