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pickupchik [31]

1 year ago

12

A carpool service has 2,000 daily riders. A one-way ticket costs $5.00. The service estimates that for each $1.00 increase to th

e one-way fare, 100 passengers will find other means of transportation. Let x represent the number of $1.00 increases in ticket price.
Choose the inequality to represent the values of x that would allow the carpool service to have revenue of at least $12,000. Then, use the inequality to select all the correct statements.

The maximum profit the company can make is $4,125.00.
The maximum profit the company can make is $15,625.00.
The price of a one-way ticket that will maximize revenue is $12.50.
The price of a one-way ticket that will maximize revenue is $7.50.

1 answer:

solmaris [256]1 year ago

7 0

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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