A bus service carries 10 000 people daily between Ajax and Unoin Station, and the company has space to serve up to 15 000 people per day. The cost to ride the bus is 20$. Market research shows that if the fare increases by $0.50, 200 fewer people will ride the bus. What fare should be charged to get the maximum revenue, given that the bus company must havve at lease $130 000 in fares a day to cover operating costs?
This question I HAVE NO IDEA how to do, can you please show me step by step so it will be easier to understand for me :) THANKYOU SOOO MUCH!
let the number of 50 cents increases be n
NOW:
number of riders = 10000
cost per ride = 20
after increase
number of riders = 10000- 200n
cost per ride = 20+ .5n
Revenue = (10000-200)(20 + .5n)
expand this and you will have a quadratic representing a parabola that opens downwards.
So the vertex will be a maximum point representing revenue
Use whatever method you learned to find the vertex
Sure, let's break down the problem step by step:
1. Determine the current number of people riding the bus:
- The bus service carries 10,000 people daily.
2. Determine the maximum capacity of the bus service:
- The company has space to serve up to 15,000 people per day.
3. Calculate the current revenue:
- The cost to ride the bus is $20.
- So, the current revenue is 10,000 people * $20 = $200,000.
4. Determine the number of people who will ride the bus if the fare increases by $0.50:
- Market research indicates that the fare increase will result in 200 fewer people riding the bus.
- So, the new number of people riding the bus will be 10,000 - 200 = 9,800.
5. Calculate the new revenue with the higher fare:
- The increased fare is $20 + $0.50 = $20.50.
- So, the new revenue is 9,800 people * $20.50 = $200,900.
6. Determine the fare that will generate maximum revenue:
- We need to find the fare that maximizes revenue while still ensuring at least $130,000 in fare revenue.
- Let's assume the fare increase is x dollars.
- The new fare will be $20 + x.
- The new number of people riding the bus will be 10,000 - (200 * x).
- The new revenue will be (10,000 - (200 * x)) * ($20 + x).
7. Create an equation to solve for x:
- We want to find the value of x that maximizes the revenue, so we need to maximize the revenue equation:
Revenue = (10,000 - (200 * x)) * ($20 + x).
- However, we also need to ensure that the revenue is at least $130,000, so we set up another equation:
(10,000 - (200 * x)) * ($20 + x) >= $130,000.
8. Solve the equation:
- Simplify the equation and solve for x.
9. Calculate the new fare and revenue with the optimized fare:
- Plug in the value of x into the equation to find the optimized fare.
- Calculate the corresponding revenue using the optimized fare and the new number of people riding the bus.
10. Compare the new revenue with the operating cost requirement:
- Check if the new revenue covers the operating cost of at least $130,000.
- If it does, then the optimized fare is the answer. If it doesn't, adjust the fare accordingly and repeat steps 6-10 until the operating cost requirement is met.
Feel free to let me know if you have any specific values or variables, and I can help you solve the equation in step 8.
Sure! Let's break down the problem step by step:
Step 1: Calculate the current revenue:
Currently, the bus service carries 10,000 people daily. The fare is $20 per person. Therefore, the current revenue is:
Current revenue = Number of people * Fare = 10,000 * $20
Step 2: Calculate the revenue if the fare increases:
If the fare increases by $0.50, 200 fewer people will ride the bus. So, the new number of people riding the bus would be 10,000 - 200 = 9,800. The new fare would be $20 + $0.50 = $20.50 per person. Therefore, the new revenue is:
New revenue = Number of people * Fare = 9,800 * $20.50
Step 3: Determine the break-even fare:
To cover the operating costs of at least $130,000, the revenue should be equal to or greater than $130,000. Therefore, we can calculate the break-even fare required to achieve the minimum revenue:
Break-even fare = Operating costs / Number of people = $130,000 / 10,000
Step 4: Compare revenues:
Now, let's compare the current revenue, new revenue, and the break-even revenue to determine the maximum revenue:
a) If the current revenue is greater than the break-even revenue, then the current fare is the one that will maximize revenue.
b) If the new revenue is greater than the current revenue and the break-even revenue, then the new fare will maximize revenue.
c) If the break-even revenue is greater than both the current and new revenue, then the break-even fare is the one that will maximize revenue.
You can calculate each of these revenues and check which condition is satisfied.
I hope this helps! Let me know if you have any further questions.
To find the fare that will generate maximum revenue, we need to analyze the relationship between the fare and the number of people riding the bus, as well as the total revenue.
Let's break it down step by step:
Step 1: Define the variables.
Let's denote:
- "F" as the fare in dollars
- "R" as the number of people riding the bus
- "TR" as the total revenue in dollars
Step 2: Understand the given information.
From the problem statement, we know that:
- The bus company carries 10,000 people daily, and it has space for up to 15,000 people.
- If the fare increases by $0.50, 200 fewer people will ride the bus.
- The operating costs must be covered by generating at least $130,000 in fares per day.
Step 3: Express the relationship between fare and the number of people.
We know that if the fare increases by $0.50, 200 fewer people will ride the bus. This can be expressed mathematically as:
R = 10,000 - (F - 20) / 0.5,
where F is the fare. This equation models the relationship between the fare and the number of people riding the bus.
Step 4: Express the total revenue.
The total revenue is calculated by multiplying the fare by the number of people riding the bus, as:
TR = R * F.
Step 5: Express the condition for operating costs.
The operating costs must be covered by generating at least $130,000 in fares per day. This can be expressed as an inequality:
TR ≥ 130,000
Step 6: Update the total revenue equation.
Replace the expression for R in the total revenue equation:
TR = (10,000 - (F - 20) / 0.5) * F
Step 7: Maximize the total revenue.
To find the fare that maximizes the total revenue, we differentiate the total revenue equation with respect to F and find its maximum point. In this case, we calculate the derivative and solve for F when the derivative equals zero.
d(TR) / dF = 0
Step 8: Simplify the derivative equation.
Differentiating the total revenue equation and setting it to zero gives the following equation:
d(TR) / dF = (20,000 - 2F) / 0.5 = 0
Step 9: Solve for F.
Solving the equation:
20,000 - 2F = 0
-2F = -20,000
F = 10,000
Step 10: Check the condition for operating costs.
Now that we have obtained a value for F, we can substitute it back into the total revenue equation to calculate the corresponding TR:
TR = (10,000 - (10,000 - 20) / 0.5) * 10,000 = 130,000
Step 11: Analyze the results.
The fare that maximizes revenue is $10. At this fare, 10,000 people will ride the bus, resulting in a total revenue of $130,000. This is sufficient to cover the operating costs of $130,000 per day.
Therefore, the fare that should be charged to get maximum revenue, while meeting the operating costs, is $10.