Answer:
Total number of riders that ride on carpool daily = 2000
Total Cost of one way ticket = $ 5.00
Total Amount earned if 2000 passengers rides daily on carpool = 2000 × 5
= $10,000
If fare increases by $ 1.00
New fare = $5 + $1
= $6
Number of passengers riding on carpool = 2,000 - 100 = 1,900
If 1,900 passengers rides on carpool daily , total amount earned ,if cost of each ticket is $ 6 = 1900 × $6 = $11400
As we have to find the inequality which represents the values of x that would allow the carpool service to have revenue of at least $12,000.
For $ 1 increase in fare = (2,000 - 1 × 100) passengers
For $ x increase in fare, number of passengers = 2,000 - 100·x
= (2,000 - 100·x) passengers
New fare = 5 + x
New Fare × Final Number of passengers ≥ 12,000
(5+x)·(2,000 - 100 x) ≥ 12,000
5 (2,000 - 100 x) + x(2,000 - 100 x) ≥ 12,000
10,000 - 500 x + 2,000 x - 100 x² ≥ 12,000
100 - 5 x + 20 x - x² ≥ 120
- x² + 15 x +100 - 120 ≥ 0
-x² + 15 x -20 ≥ 0
x² - 15 x + 20 ≤ 0
⇒ x = 1.495
x ≥ $ 1.495, that is if we increase the fare by this amount or more than this the revenue will be at least 12,000 or more .
Also, f'(x) = 0 gives x = 7.5
⇒ The price of a one-way ticket that will maximize revenue is $7.50
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