A community college campus plans to construct a rectangular parking lot on land bordered on one side by a highway. There are 640 feet of fencing available to fend the other three sides. Let x represent the length of each of the two parallel sides of fencing.

a) Express the length of the remaining side to be fenced in terms of x. b) What are the restrictions on x? c) Determine the function A that represents the area of the parking lot in terms of x. d) Determine the values of x that will give an area between 30,000 and 40,000 square feet. e) What dimensions will give a maximum area, and what will this area be?

a) Since there are two parallel sides of length x, the remaining side to be fenced will have a length of 640 - 2x.

b) The restrictions on x are that it must be greater than zero (x > 0) and less than half the length of the available fencing (x < 320).

c) The area of the rectangular parking lot can be calculated by multiplying the length and width, which gives A = x * (640 - 2x).

d) To find the values of x that will give an area between 30,000 and 40,000 square feet, we set up the following inequality: 30,000 < x * (640 - 2x) < 40,000. Solving this inequality will give the range of values for x.

e) To find the dimensions that will give a maximum area, we need to maximize the function A = x * (640 - 2x). We can do this by finding the critical points of the function and evaluating them.

a) Let the length of each of the two parallel sides of fencing be x. Since the parking lot is rectangular, the remaining side is also x.

b) The restrictions on x are that it must be positive (since sides cannot have negative lengths) and the total length of the three sides cannot exceed the available 640 feet of fencing.

c) The area A of the parking lot can be calculated by multiplying the length and width of the rectangle. The length is x and the width is also x, so the area function A in terms of x is: A(x) = x * x = x^2.

d) To find the values of x that will give an area between 30,000 and 40,000 square feet, we can set up the following inequality:
30,000 <= x^2 <= 40,000

First, let's solve the lower bound of the inequality:
30,000 <= x^2
Taking the square root of both sides:
sqrt(30,000) <= x
x >= sqrt(30,000)

Next, let's solve the upper bound of the inequality:
x^2 <= 40,000
Taking the square root of both sides:
x <= sqrt(40,000)

Therefore, the values of x that will give an area between 30,000 and 40,000 square feet are x >= sqrt(30,000) and x <= sqrt(40,000).

e) To find the dimensions that will give a maximum area, we need to find the turning point (maximum) of the quadratic function A(x) = x^2. Since the coefficient of x^2 is positive, the parabola opens upward. The x-coordinate of the turning point can be found using the formula: x = -b/2a, where a = 1 and b = 0.

Therefore, x = -0/2(1) = 0. The length and width of the parking lot cannot be zero, so we find the area for values of x close to zero:

When x is very close to zero, such as 0.0001, the area is approximately 0.0001^2 = 0.00000001 square feet.

The dimensions that will give a maximum area are x = 0, and the maximum area is 0 square feet.

a)

Total: 640 feet
Two short sides: 2x
long side 640-2x
b)
Since long side must be positive, we have 640-2x≥0
c)
Area = short side * long side
A(x)=x(640-2x)
d)
find x such that
30000≤A(x)≤40000
Make a table that gives x and A(x),
Hint:
A(50)=27000 <30000
A(86)=40248 >40000

e)
If you have done calculus, find A'(x) and equate to zero.
If you have not, then complete squares of A(x):
A(x)=ax²+bx+c
-2x²+640x
where a=-2, b=640, c=0
The maximum occurs at x=-b/2a