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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.
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