Question 183949
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The perimeter of a rectangle is given by:


*[tex \LARGE \text{          }\math P = 2l + 2w \ \ \Rightarrow\ \ 2l = P - 2w \ \ \Rightarrow\ \ l = \frac{P}{2} - w]


The area of a rectangle is given by:


*[tex \LARGE \text{          }\math A = lw]


Substituting from the perimeter equation:


*[tex \LARGE \text{          }\math A(w) = w\left(\frac{P}{2} - w\right)\ \ \Rightarrow\ \ A(w) = \left(\frac{wP}{2} - w^2\right)]


This function graphs to a parabola opening downward meaning that the vertex is a maximum.  The maximum value of the function, hence the maximum area, is where the value of the first derivative is equal to zero:



*[tex \LARGE \text{          }\math \frac{d}{dw}A(w) = A'(w) = -2w + \frac{P}{2}]


Set equal to zero:


*[tex \LARGE \text{          }\math -2w + \frac{P}{2} = 0 \ \ \Rightarrow\ \ 2w = \frac{P}{2} \ \ \Rightarrow\ \ w = \frac{P}{4}]


Hence, the maximum area rectangle for a given perimeter is a square with sides of length one-fourth of the perimeter.


{{{drawing(
500, 500, -5, 16, -10, 60,
grid(1),
circle(7.5,(7.5)^2,.1),
graph(
500, 500, -5, 16, -10, 60,
y = -x^2 + 15x
))}}}


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