Question

A tour operator charges Rs. 200 per passenger for 50 passengers with a discount of

Rs. 5 for each 10 passenger in excess of 50. Determine the number of passengers that

will maximize the revenue of the operator​

Answer 1

Given :  A tour operator charge rs 200 per passenger for 50 passenger with a discount of rs 5 for each 10 passenger in excess of 50.

To find : The number of passengers that will maximize the revenue of the operator​.

Solution:

• Now we have given that operator charge Rs 200 per passenger for 50 passenger, so revenue will be:

                 = 200 x 50  

                 = Rs 10000

• Now discount of Rs 5 for each 10 passenger in excess of 50 will be:

• Consider 10x  Passenger in excess of 50.

• So the charges will be: 200  - 5x

• Now revenue is:

                 = (200 - 5x)(50 + 10x)

                 = 10000 + 2000x - 250x  - 50x²

                 = 10000 + 1750x - 50x²

• Now differentiating the revenue with respect to x, we get:

                 dR/dx =  1750  - 100x

                 1750 - 100P = 0

                 x = 17.5

                 d²R/dx² =   - 100  

                 d²R/dx² < 0

• So the maximum revenue at x is 17.5

• But x should be integer, so consider 17 and 18, we get:

• Revenue = (200 - 5x)(50 + 10x)

• If x = 17

• Then Revenue will be:

                 =  (200 - 85) (50 + 170)  

                 =  25300

• If x = 18

• Then Revenue will be:

                 =  (200 - 90) (50 + 180)  

                 =  25300

• So the passengers  will be 220  or 230 will maximize the revenue of the operator​.

• And if we consider 17.5, then Revenue will be:

                 =  (200 - 87.5) (50 + 175)  

                 =  25312.5

                 Passengers = 225

Answer:

         So the number of passengers will be  220 to 230.