Question
The figure shows costs for rectangular lots of different widths. Each lot is twice as long as it is wide. To coat a parking lot 300 meters long and 150 meters wide, a developer budgeted $20,250, or three times the cost of a lot 50 meters wide. Will the budget be sufficient? Justify your answer using a function model.
Cost of Reflective Coating
250
1090
2450
4300
6750
Width of Lot (meters)
10
20
30
40
50
Produce a function model for the data. Let x be the width and f(x) be the cost to coat a parking lot of that width.
Cost of Reflective Coating
250
1090
2450
4300
6750
Width of Lot (meters)
10
20
30
40
50
Produce a function model for the data. Let x be the width and f(x) be the cost to coat a parking lot of that width.
Answers
GPT-4o mini
To analyze this problem, we need to create a function model based on the costs associated with different widths of lots as provided in the data. We'll first outline the data given:
### Given Data Points:
- At \(x = 10\) meters, cost \(C(10) = \$250\)
- At \(x = 20\) meters, cost \(C(20) = \$1090\)
- At \(x = 30\) meters, cost \(C(30) = \$2450\)
- At \(x = 40\) meters, cost \(C(40) = \$4300\)
- At \(x = 50\) meters, cost \(C(50) = \$6750\)
### Suggested Function Model:
From the data points, we can observe that the cost of coating is likely a polynomial function of the width, perhaps a quadratic function. To create a function model, we can use polynomial regression or fit a quadratic function to the given data points.
To find a quadratic function, we can express it in the form:
\[
f(x) = ax^2 + bx + c
\]
Since we have five data points, we can solve for the coefficients \(a\), \(b\), and \(c\) using a system of equations derived from plugging in the x-values and their corresponding costs.
### Using the Data to Fit the Quadratic Model:
We can create a system of equations based on the given values.
1. From \( f(10) = 250 \):
\[
100a + 10b + c = 250 \quad (1)
\]
2. From \( f(20) = 1090 \):
\[
400a + 20b + c = 1090 \quad (2)
\]
3. From \( f(30) = 2450 \):
\[
900a + 30b + c = 2450 \quad (3)
\]
We can create equations by eliminating \(c\) using equations (1), (2), and (3):
Subtract (1) from (2):
\[
(400 - 100)a + (20 - 10)b + (c - c) = 1090 - 250 \\
300a + 10b = 840 \quad (4)
\]
Subtract (2) from (3):
\[
(900 - 400)a + (30 - 20)b + (c - c) = 2450 - 1090 \\
500a + 10b = 1360 \quad (5)
\]
Now, we can simplify (4) and (5):
From (4):
\[
30a + b = 84 \quad (6)
\]
From (5):
\[
50a + b = 136 \quad (7)
\]
Now subtract (6) from (7):
\[
(50 - 30)a + (b - b) = 136 - 84 \\
20a = 52 \\
a = 2.6
\]
Then substitute \(a\) into (6):
\[
30(2.6) + b = 84 \\
78 + b = 84 \\
b = 6
\]
Finally, substitute \(a\) and \(b\) back into (1) to solve for \(c\):
\[
100(2.6) + 10(6) + c = 250 \\
260 + 60 + c = 250 \\
c = 250 - 320 \\
c = -70
\]
### Final Model:
Thus, our quadratic function model for the cost of coating a lot based on its width is:
\[
f(x) = 2.6x^2 + 6x - 70
\]
### Cost of Coating the Developer's Lot:
Given the lot is 150 meters wide (which is \(x = 150\)), we will calculate the cost:
\[
f(150) = 2.6(150^2) + 6(150) - 70
\]
\[
= 2.6(22500) + 900 - 70
\]
\[
= 58500 + 900 - 70
\]
\[
= 58500 + 830
\]
\[
= 59330
\]
### Comparing with Budget:
The developer's budget is $20,250. Since $59,330 exceeds the budget of $20,250, we can conclude that the budget will **not be sufficient** to cover the cost of coating the parking lot.
### Given Data Points:
- At \(x = 10\) meters, cost \(C(10) = \$250\)
- At \(x = 20\) meters, cost \(C(20) = \$1090\)
- At \(x = 30\) meters, cost \(C(30) = \$2450\)
- At \(x = 40\) meters, cost \(C(40) = \$4300\)
- At \(x = 50\) meters, cost \(C(50) = \$6750\)
### Suggested Function Model:
From the data points, we can observe that the cost of coating is likely a polynomial function of the width, perhaps a quadratic function. To create a function model, we can use polynomial regression or fit a quadratic function to the given data points.
To find a quadratic function, we can express it in the form:
\[
f(x) = ax^2 + bx + c
\]
Since we have five data points, we can solve for the coefficients \(a\), \(b\), and \(c\) using a system of equations derived from plugging in the x-values and their corresponding costs.
### Using the Data to Fit the Quadratic Model:
We can create a system of equations based on the given values.
1. From \( f(10) = 250 \):
\[
100a + 10b + c = 250 \quad (1)
\]
2. From \( f(20) = 1090 \):
\[
400a + 20b + c = 1090 \quad (2)
\]
3. From \( f(30) = 2450 \):
\[
900a + 30b + c = 2450 \quad (3)
\]
We can create equations by eliminating \(c\) using equations (1), (2), and (3):
Subtract (1) from (2):
\[
(400 - 100)a + (20 - 10)b + (c - c) = 1090 - 250 \\
300a + 10b = 840 \quad (4)
\]
Subtract (2) from (3):
\[
(900 - 400)a + (30 - 20)b + (c - c) = 2450 - 1090 \\
500a + 10b = 1360 \quad (5)
\]
Now, we can simplify (4) and (5):
From (4):
\[
30a + b = 84 \quad (6)
\]
From (5):
\[
50a + b = 136 \quad (7)
\]
Now subtract (6) from (7):
\[
(50 - 30)a + (b - b) = 136 - 84 \\
20a = 52 \\
a = 2.6
\]
Then substitute \(a\) into (6):
\[
30(2.6) + b = 84 \\
78 + b = 84 \\
b = 6
\]
Finally, substitute \(a\) and \(b\) back into (1) to solve for \(c\):
\[
100(2.6) + 10(6) + c = 250 \\
260 + 60 + c = 250 \\
c = 250 - 320 \\
c = -70
\]
### Final Model:
Thus, our quadratic function model for the cost of coating a lot based on its width is:
\[
f(x) = 2.6x^2 + 6x - 70
\]
### Cost of Coating the Developer's Lot:
Given the lot is 150 meters wide (which is \(x = 150\)), we will calculate the cost:
\[
f(150) = 2.6(150^2) + 6(150) - 70
\]
\[
= 2.6(22500) + 900 - 70
\]
\[
= 58500 + 900 - 70
\]
\[
= 58500 + 830
\]
\[
= 59330
\]
### Comparing with Budget:
The developer's budget is $20,250. Since $59,330 exceeds the budget of $20,250, we can conclude that the budget will **not be sufficient** to cover the cost of coating the parking lot.