SOLUTION: A rectangular parking lot must have a perimeter of 520 feet and an area of at least 12,000 square feet. Describe the possible lengths of the parking lot.

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Question 1204973: A rectangular parking lot must have a perimeter of 520 feet and an area of at least 12,000 square feet. Describe the possible lengths of the parking lot.
Found 3 solutions by josgarithmetic, math_tutor2020, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!

No negative values allowed.
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Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

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

plug in y = -x+260

Consider the equation

Use the quadratic formula to solve and you'll find the x intercepts are x = 60 and x = 200

If , then f(x) = -x^2+260x-12000 is nonnegative and leads back to being true.

is only true when

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 to find its paired y value.
Make sure the x and y values add to 260.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

Here is an easy way to find the answer.

The perimeter is 520 feet, and the area has to be at least 12000 square feet.

The maximum area with a given perimeter is if the rectangle is a square; the side length of the square would be 130 feet. So

Let 130+x be the length of the parking lot
then 130-x is the width of the parking lot

The area is greater than or equal to 12000:

To have an area of 12000 square feet or more, the length can be 130+x, where the maximum value of x is 70 (and, or course, the minimum value is 0). That means the length can be anywhere from 130 feet to 130+70 = 200 feet.

ANSWER: The possible lengths are from 130 to 200 feet