Question 992341
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Using the formula for the perimeter of a rectangle, it can be easily shown that the sum of the length and the width of a rectangle is equal to one-half of the perimeter, which is to say:


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


Or, in the case of this problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 55\ -\ w]


Since the area is the length times the width,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ 55w\ -\ w^2]


gives the area of any rectangle that can be formed with a perimeter of 110 as a function of the width of the rectangle.


We seek the width, from which we can determine the length, of a rectangle with a perimeter of 110 feet, such that the area is greater than or equal to 700 square feet.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -w^2\ +\ 55w\ -\ 700\ \geq\ 0]


Find the zeros of the corresponding quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -w^2\ +\ 55w\ -\ 700\ =\ 0]


Since the area function given above is a quadratic function with a negative lead coefficient, the graph is a parabola that opens downward.  Hence, the portion of the graph that lies between the two zeros of the function, inclusive is the portion of the quadratic inequality that is greater than zero.  So the zeros are the limits of the closed interval where the area is 700 square feet or greater.


     *[illustration AreaMoreThan700.jpg].
 

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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \