One campus of HCC has plans to construct rectangular parking lot on land bordered on one side by a highway. There are 640 ft of fencing available to fence the other sides. Let x represent the length of each of the two parallel sides of fencing.
a. draw the illustration of the problem
b.express length of remaining side to be fenced in terms of x.
c. what're the restrictions on x?
d. Determine values of x that'll give area between 30,000 and 40,000 ft squared.
e. what dimensions will give a max area, and what will this area be?
f. Determine a function A that represents the area of the parking lot in terms of x.
a) --- up to you
let the length of the side which is parallel to the highway be x , x < 640
let the length of the other two equal sides by y each
x + 2y = 640 ---> y = (640-x)/2 = 320 - x/2
area = xy = x(320-x/2)
= 320x - (1/2)x^2
so we need the vertex of this downwards opening parabola
the x of the vertex = -b/(2a) = -320/(-1) = 320
if x = 320, then y = (640-320)/2 = 160
and the maximum area is (160)(320) = 51200 ft^2
for part d)
30000 = 320x - x^2/2
x^2 - 640x + 60000 = 0
x = (640 ± √169600)/2 = 114.1 ft or 529 ft
if x = 114.1 then y = 262.95 and area = 30003
if x = 529 ft then y = negative, no way
40,000 = 320x - x^2/2
x^2 -640x + 80000 = 0
x = 170.33 , then y = 234.83 ft, area = 39998.6
I will leave it up to you to split up my solutions into the individual parts b), c) etc
a. In order to visualize the problem, let's draw a diagram.
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| Parking Lot |
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Highway
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b. Let's express the length of the remaining side to be fenced in terms of x.
Since the parking lot is rectangular, it has two pairs of parallel sides. The side that is not bordered by the highway is divided into two sections, each of length x. Therefore, the length of the remaining side that needs to be fenced is also x.
c. Now let's talk about the restrictions on x.
The total length of fencing available is 640 ft. This means that the sum of the lengths of all four sides should equal 640 ft.
So, we can write an equation: 2x + x + x = 640. Simplifying, we get 4x = 640, and dividing both sides by 4, we find x = 160.
d. To determine the values of x that give an area between 30,000 and 40,000 square feet, we can use the formula for the area of a rectangle:
Area = length * width
In this case, the length is x and the width is also x.
So, Area = x * x = x^2.
We can now substitute different values for x (within the given range) and calculate the corresponding areas. The values of x that result in an area between 30,000 and 40,000 square feet are the solutions.
e. To find the dimensions that give the maximum area, we need to differentiate the area function and set its derivative equal to zero. However, since we don't have an explicit function for the area in terms of x, we can't find the maximum area directly.
However, we can observe that a square will have the maximum area for a given perimeter in a rectangle. Therefore, in this case, the dimensions that give the maximum area will be a square shape.
So, for the parking lot, the dimensions that give the maximum area will be x = 160 ft (length) and x = 160 ft (width), resulting in a square shape.
The area will be: Area = x * x = 160 ft * 160 ft = 25,600 square feet.
f. To determine a function A that represents the area of the parking lot in terms of x, we can use the formula for the area of a rectangle:
A = length * width.
Since both the length and width of the parking lot are x, we can substitute this into the formula:
A = x * x = x^2.
So, the function A that represents the area of the parking lot in terms of x is A(x) = x^2.