station carries 15000 bus riders per day for a fare of 1.90. the city wishes to increase ridership in an effort to reduce car polltuion, while at the same time maximizing the transit revenues. tehe city conducted a survey and the results indicated that number of riders would increase by 500 for every 0.05$ decrease in fare.
1-write an equation to represent daily revenue
2-what fare will produce the greatest revenue? what is the expected daily rideship?
3city thinks 24675 per day to break even. what tickets prices would enable the city to earn a profit?
1. To write an equation to represent daily revenue, we need to determine the relationship between the fare and the number of riders.
Let's denote:
R = Daily revenue
F = Fare per ticket
N = Number of riders
Here is the equation to represent daily revenue:
R = F * N
2. To find the fare that will produce the greatest revenue and the expected daily ridership, we can use the information given in the survey. According to the survey results, the number of riders increases by 500 for every $0.05 decrease in fare.
Let's denote:
N₀ = Initial number of riders (15,000)
ΔF = Decrease in fare (in dollars)
ΔN = Increase in the number of riders
From the survey results, we have the relationship:
ΔN = 500 * (ΔF / 0.05)
To find the fare that produces the greatest revenue, we need to maximize the revenue equation R (from step 1) with respect to the fare F. Since daily revenue is proportional to the number of riders, maximizing ridership will maximize revenue.
Maximizing ΔN will result in maximizing ridership, and it can be achieved by maximizing the decrease in fare ΔF (assuming the fare doesn't go below $0).
3. If the city wants to break even at 24,675 riders per day, we need to find the ticket prices that will enable the city to earn a profit.
Let's denote:
N₁ = Target number of riders for break-even (24,675)
Assuming all costs per rider are covered by the fare price, the equation for daily revenue is:
R = F * N
To break even, the revenue should equal the cost:
R = Cost
Given that the daily revenue (R) is equal to the ticket price (F) multiplied by the number of riders (N), we have:
F * N = Cost
Thus, we can solve for the fare price (F) that will allow the city to earn a profit by rearranging the equation:
F = Cost / N₁
By substituting the given break-even number of riders into the equation, we can determine the ticket prices that will enable the city to earn a profit.
1. To write an equation to represent the daily revenue, we need to consider the fare and the number of daily riders. Let's break it down step by step:
Let:
- x be the decrease in fare in dollars ($)
- R be the daily revenue in dollars ($)
- N be the number of daily riders
Given:
- Current fare = $1.90
- Current number of riders per day = 15,000
- Increase in riders for every $0.05 decrease in fare = 500
First, we need to determine the new number of riders based on the fare decrease. Since there is an increase of 500 riders for every $0.05 decrease, we can represent this relationship as follows:
New number of riders = Current number of riders + (500 * (Current fare - New fare) / 0.05)
Substituting in the given values:
New number of riders = 15,000 + (500 * (1.90 - (1.90 - x)) / 0.05)
Simplifying the equation:
New number of riders = 15,000 + (500 * x / 0.05)
Now, we can determine the daily revenue using the fare and the number of riders:
Daily revenue = Fare x Number of riders
Daily revenue = (1.90 - x) * New number of riders
Daily revenue = (1.90 - x) * (15,000 + (500 * x / 0.05))
Therefore, the equation to represent daily revenue is:
R = (1.90 - x) * (15,000 + (500 * x / 0.05))
2. To find the fare that will produce the greatest revenue and the expected daily ridership, we need to determine the maximum value of the daily revenue equation:
Since the equation is quadratic, we can find the maximum by taking the derivative of R with respect to x, setting it equal to zero, and solving for x.
dR/dx = 0
(1.90 - x) * (15,000 + (500 * x / 0.05)) = 0
Solving for x:
1.90 - x = 0 (Based on the first factor) OR 15,000 + (500 * x / 0.05) = 0 (Based on the second factor)
If we solve for x using the first factor, we get:
1.90 - x = 0
x = 1.90
If we solve for x using the second factor, we get:
15,000 + (500 * x / 0.05) = 0
500 * x / 0.05 = -15,000
500 * x = -15,000 * 0.05
x = -150
Since a negative fare does not make sense in this context, we discard the second solution.
Therefore, the fare that will produce the greatest revenue is $1.90, and the expected daily ridership would be 15,000.
3. To determine the ticket prices that would enable the city to earn a profit of $24,675 per day, we can set the daily revenue equation equal to the break-even revenue and solve for x:
R = (1.90 - x) * (15,000 + (500 * x / 0.05))
24,675 = (1.90 - x) * (15,000 + (500 * x / 0.05))
Simplifying and solving for x:
(1.90 - x) * (15,000 + (500 * x / 0.05)) = 24,675
(1.90 - x) * (15,000 * 0.05 + 500 * x) = 24,675
We can solve this equation to find the appropriate ticket prices that would enable the city to earn a profit.