a City's transit authority serves 178500 communters daily when the fare is $1.90. Market research has determined that every penny decrease in the fare will result in 1,050 new riders. what fare will maximize revenue?

revenue = #riders * fare

= (178500+1050(190-x))(.01x) for x<=190

that's just a parabola with vertex at x=180.

number of penny decreases ---- n

cost of fare = 190-n
number of riders = 178500 + 1050n

revenue = R = (190-n)(178500+1050n)
P' = (190-n)(1050) + (178500 + 1050n)(-1)
= 0 for a max of P

178500 + 1050n = 1050(190-n)
divide by 1050
170 + n = 190-n
2n = 20
n = 10

the fare should be 190+10 = 200
or it should be $ 2.00

To find the fare that will maximize revenue, we need to determine the point at which the increase in ridership due to a decrease in fare is balanced with the decrease in revenue per rider. Let's break down the steps:

1. Start with the given information:
- Current fare: $1.90
- Number of daily commuters at the current fare: 178,500
- Increase in riders per penny decrease in fare: 1,050

2. Determine the relationship between fare and ridership:
- For every penny decrease in fare, there will be an increase of 1,050 new riders.
- This means that for every penny increase in fare, there will be a decrease of 1,050 riders.

3. Calculate the decrease in ridership for each $0.01 increase in fare:
- Since 1 penny = $0.01, we divide 1,050 by 100 to find the decrease in ridership per $0.01 increase in fare:
1,050 new riders / 100 = 10.5 riders per $0.01 increase in fare.

4. Determine the fare that maximizes revenue:
- Start with the current fare of $1.90 and calculate the decrease in fare required to maintain or increase ridership.

Fare decrease = (Ridership decrease / Increase in riders per penny decrease in fare) * $0.01
Fare decrease = (10.5 riders / 1 rider per $0.01) * $0.01
Fare decrease = $0.105

- Subtract the fare decrease from the current fare to find the fare that maximizes revenue:

Fare = Current fare - Fare decrease
Fare = $1.90 - $0.105
Fare = $1.795

Therefore, a fare of $1.795 will maximize revenue for the transit authority.