Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Ask your questions and receive precise answers from experienced professionals across different disciplines. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Let's break down the problem step-by-step and address each part of it.
### Step 1: Creating the Revenue Function
First, we define the total revenue \( R \) with respect to \( x \), where \( x \) is the number of \$1.00 increases to the ticket price:
- Initial number of daily riders: 2000
- Decrease in riders per \$1.00 increase: 100
- Initial ticket cost: \$5.00
- Revenue target: \$12,000
Thus, we express the revenue function as:
[tex]\[ R(x) = (\text{initial riders} - \text{decrease per increase} \cdot x) \cdot (\text{initial cost} + x) \][/tex]
[tex]\[ R(x) = (2000 - 100x) \cdot (5 + x) \][/tex]
### Step 2: Setting Up the Inequality
We want the revenue to be at least \$12,000. Therefore, we set up the inequality:
[tex]\[ (2000 - 100x)(5 + x) \geq 12000 \][/tex]
### Step 3: Simplifying the Inequality
To solve for \( x \), we first expand and simplify the inequality:
[tex]\[ (2000 - 100x)(5 + x) \geq 12000 \][/tex]
Expanding the left-hand side:
[tex]\[ 100x^2 - 1500x - 10000 \geq 12000 \][/tex]
Rewriting it in standard quadratic form gives us:
[tex]\[ 100x^2 - 1500x - 22000 \geq 0 \][/tex]
### Step 4: Solving the Quadratic Equation
Next, we solve the quadratic equation \( 100x^2 - 1500x - 22000 = 0 \):
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values \( a = 100 \), \( b = -1500 \), and \( c = -22000 \):
[tex]\[ \Delta = b^2 - 4ac = (-1500)^2 - 4 \cdot 100 \cdot (-22000) = 11050000 \][/tex]
The roots are:
[tex]\[ x = \frac{1500 \pm \sqrt{11050000}}{200} \][/tex]
This yields two solutions:
[tex]\[ x_1 = 24.12 \][/tex]
[tex]\[ x_2 = -9.12 \][/tex]
Since \( x \) represents the number of \$1.00 increases, it must be non-negative. Therefore, we discard \( x = -9.12 \) as it is not feasible, leaving \( x = 24.12 \).
### Step 5: Maximum Profit and Ticket Price for Maximum Revenue
To find the maximum revenue, we use the vertex form of the quadratic equation. The vertex \( x = -\frac{b}{2a} \) for the quadratic function \( 100x^2 - 1500x - 22000 \):
[tex]\[ x = \frac{1500}{2 \cdot 100} = 7.5 \][/tex]
Plugging \( x = 7.5 \) into the revenue function:
[tex]\[ R(7.5) = (2000 - 100 \cdot 7.5)(5 + 7.5) = 1250 \times 12.5 = 15625 \][/tex]
### Step 6: Summary of Results
Based on the calculations:
1. The correct inequality to represent the values of \( x \) that would allow the carpool service to have revenue of at least \$12,000 is:
[tex]\[ 100x^2 - 1500x - 22000 \geq 0 \][/tex]
2. The maximum profit the company can make is \( \$15625.00 \).
3. The price of a one-way ticket that will maximize revenue is \( \$7.50 \).
### Incorrect Statements
The statements suggesting a maximum profit of \[tex]$4125.00 and a one-way ticket price of \$[/tex]12.50 for maximum revenue are incorrect.
### Final Selection of Correct Statements
- The maximum profit the company can make is \$15,625.00.
- The price of a one-way ticket that will maximize revenue is \$7.50.
These are the correct conclusions drawn from the problem statement and the subsequent mathematical analysis.
### Step 1: Creating the Revenue Function
First, we define the total revenue \( R \) with respect to \( x \), where \( x \) is the number of \$1.00 increases to the ticket price:
- Initial number of daily riders: 2000
- Decrease in riders per \$1.00 increase: 100
- Initial ticket cost: \$5.00
- Revenue target: \$12,000
Thus, we express the revenue function as:
[tex]\[ R(x) = (\text{initial riders} - \text{decrease per increase} \cdot x) \cdot (\text{initial cost} + x) \][/tex]
[tex]\[ R(x) = (2000 - 100x) \cdot (5 + x) \][/tex]
### Step 2: Setting Up the Inequality
We want the revenue to be at least \$12,000. Therefore, we set up the inequality:
[tex]\[ (2000 - 100x)(5 + x) \geq 12000 \][/tex]
### Step 3: Simplifying the Inequality
To solve for \( x \), we first expand and simplify the inequality:
[tex]\[ (2000 - 100x)(5 + x) \geq 12000 \][/tex]
Expanding the left-hand side:
[tex]\[ 100x^2 - 1500x - 10000 \geq 12000 \][/tex]
Rewriting it in standard quadratic form gives us:
[tex]\[ 100x^2 - 1500x - 22000 \geq 0 \][/tex]
### Step 4: Solving the Quadratic Equation
Next, we solve the quadratic equation \( 100x^2 - 1500x - 22000 = 0 \):
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values \( a = 100 \), \( b = -1500 \), and \( c = -22000 \):
[tex]\[ \Delta = b^2 - 4ac = (-1500)^2 - 4 \cdot 100 \cdot (-22000) = 11050000 \][/tex]
The roots are:
[tex]\[ x = \frac{1500 \pm \sqrt{11050000}}{200} \][/tex]
This yields two solutions:
[tex]\[ x_1 = 24.12 \][/tex]
[tex]\[ x_2 = -9.12 \][/tex]
Since \( x \) represents the number of \$1.00 increases, it must be non-negative. Therefore, we discard \( x = -9.12 \) as it is not feasible, leaving \( x = 24.12 \).
### Step 5: Maximum Profit and Ticket Price for Maximum Revenue
To find the maximum revenue, we use the vertex form of the quadratic equation. The vertex \( x = -\frac{b}{2a} \) for the quadratic function \( 100x^2 - 1500x - 22000 \):
[tex]\[ x = \frac{1500}{2 \cdot 100} = 7.5 \][/tex]
Plugging \( x = 7.5 \) into the revenue function:
[tex]\[ R(7.5) = (2000 - 100 \cdot 7.5)(5 + 7.5) = 1250 \times 12.5 = 15625 \][/tex]
### Step 6: Summary of Results
Based on the calculations:
1. The correct inequality to represent the values of \( x \) that would allow the carpool service to have revenue of at least \$12,000 is:
[tex]\[ 100x^2 - 1500x - 22000 \geq 0 \][/tex]
2. The maximum profit the company can make is \( \$15625.00 \).
3. The price of a one-way ticket that will maximize revenue is \( \$7.50 \).
### Incorrect Statements
The statements suggesting a maximum profit of \[tex]$4125.00 and a one-way ticket price of \$[/tex]12.50 for maximum revenue are incorrect.
### Final Selection of Correct Statements
- The maximum profit the company can make is \$15,625.00.
- The price of a one-way ticket that will maximize revenue is \$7.50.
These are the correct conclusions drawn from the problem statement and the subsequent mathematical analysis.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
