Question 1204973
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x = length
y = width
each is in feet


2(length+width) = perimeter of rectangle
2(x+y) = perimeter of rectangle
2(x+y) = 520
x+y = 520/2
x+y = 260
y = -x+260


length*width = xy = area of the rectangle
{{{area >= 12000}}}


{{{xy >= 12000}}}


{{{x(-x+260) >= 12000}}} plug in y = -x+260


{{{-x^2+260x >= 12000}}}


{{{-x^2+260x-12000 >= 0}}}



Consider the equation
{{{-x^2+260x-12000 = 0}}}
Use the quadratic formula to solve and you'll find the x intercepts are x = 60 and x = 200


If {{{60 <= x <= 200}}}, then f(x) = -x^2+260x-12000 is nonnegative and leads back to {{{-x^2+260x-12000 >= 0}}} being true.


{{{xy >= 12000}}} is only true when {{{60 <= x <= 200}}}


Due to symmetry, the x coordinate of the vertex is located at the midpoint of those roots. So the x coordinate of the vertex happens when x = (60+200)/2 = 260/2 = 130


So,
y = -x+260
y = -130+260
y = 130
and
xy = 130*130 = 16900


The largest area possible is 16900 sq ft and it happens when we have a rectangle of dimensions x = 130 and y = 130; i.e. a square with side length 130 ft.


Another possible parking lot is when x = 80 and y = 180. It produces an area of xy = 80*180 = 14400 sq ft.


Another possible parking lot is when x = 100 and y = 160. It produces an area of xy = 100*160 = 16000 sq ft.



There are infinitely many lots we can form. Simply pick any x value in the interval {{{60 <= x <= 200}}} to find its paired y value. 
Make sure the x and y values add to 260.
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