Question 199484
<font face="Garamond" size="+2">


The perimeter of a rectangle is given by the formula:


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


Where *[tex \Large l] is the measure of the length of the rectangle and *[tex \Large w] is the measure of the width.


Solving for *[tex \Large l]:


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


The area of a rectangle is given by the formula:


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


Substituting the expression for *[tex \Large l] derived earlier, you can write a function for the area of the rectangle in terms of the width where the perimeter is a constant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(x) = \left(\frac{P - 2w}{2}\right)w = - w^2 + \frac{1}{2}Pw]


This is a quadratic function whose graph is a parabola.  Since the lead coefficient is <0, the parabola opens downward and the vertex of the parabola represents a maximum.  Since the independent variable is the width and the value of the function is the area, the coordinates of the vertex will tell us the width in terms of the perimeter that gives us the maximum area, and the value of that maximum area, again in terms of the perimeter.


A parabola represented by *[tex \Large f(x) = ax^2 + bx + c] has a vertex at the point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)]


For the area function derived above, *[tex \Large a = -1] and *[tex \Large b = \frac{P}{2}], so:


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


Hence, the maximum area is obtained when *[tex \Large w = \frac{P}{4}], which means that *[tex \Large 2w = \frac{P}{2}] and therefore *[tex \Large 2l = \frac{P}{2}] also.  Therefore, the maximum area for a given perimeter is obtained when the rectangle is actually a square with side measure one-fourth of the perimeter.


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