The campus of a college has plans to construct a rectangular parking lot on land bordered on one side by a highway. There are 720 feet of fencing available to fence 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 a 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 20,000 and 40,000 ft^2.
e) What dimensions will give a maximum area, and what will this area be?
The campus of a college has plans to construct a rectangular parking lot on land bordered on one side by a highway. There are 760
ft of fencing available to fence the other three sides. Let x represent the length of each of the two parallel sides of fencing.
A rectangle has width x. xx
(a) Express the length of the remaining side to be fenced in terms of x.
(b) What are the restrictions on x?
(c) Determine a 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 20 comma 000
and 50 comma 000
ftsquared
.
(e) What dimensions will give a maximum area, and what will this area be?
16
a) To find the length of the remaining side to be fenced, we need to subtract the combined lengths of the two parallel sides (x) from the total available fencing (720 feet). Since we have two sides of length x, the remaining side length can be expressed as:
Remaining side length = 720 - 2x
b) There are some restrictions on x to consider. Firstly, x must be less than half of the total available fencing because we have two sides of length x. So, the restriction is:
x < 720/2
x < 360
Also, the remaining side length (720 - 2x) must be positive for the parking lot to have a valid shape. So, the restriction is:
720 - 2x > 0
2x < 720
x < 360
Combining these restrictions, we have:
0 < x < 360
c) To determine the area A, we multiply the length (x) by the width (remaining side length):
A = x * (720 - 2x)
A = 720x - 2x^2
d) To find the values of x that will give an area between 20,000 and 40,000 square feet, we can set up the inequality:
20,000 ≤ 720x - 2x^2 ≤ 40,000
Simplifying this inequality, we get:
-2x^2 + 720x - 40,000 ≥ 0
To solve this quadratic inequality, we can find the x-intercepts by setting it equal to zero:
-2x^2 + 720x - 40,000 = 0
Using factoring, completing the square, or the quadratic formula, we can find the x-values that satisfy this equation. The solution will give us the range of x-values that meet the area requirements.
e) To find the dimensions that will give the maximum area, we need to optimize the area function. We can do this by finding the critical points of the function A(x), which occur when the derivative of A with respect to x equals zero.
dA/dx = 720 - 4x
Setting the derivative equal to zero and solving for x, we get:
720 - 4x = 0
4x = 720
x = 720/4
x = 180
To check if this value is a maximum, we can take the second derivative:
d^2A/dx^2 = -4
Since the second derivative is negative, it confirms that x = 180 gives the dimensions for a maximum area. Plugging this value back into the area function, we can find the maximum area A.