Math, asked by chintusaini371, 10 months ago

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.with explaination​

Answers

Answered by Agastya0606
3

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.

Answered by sonuvuce
1

The number of passengers that will maximize the revenue of the operator is 225

Step-by-step explanation:

Given

A tour operator charges Rs. 200 per passenger for 50 passengers

To find out

The number of passengers that will maximize the revenue

Solution

Let us assume that there are x passengers

Passengers in excess of 50 = x - 50

For every 10 passengers in excess of 50 he charges 5 Rs. less

Thus

R=(x)[200-\frac{5}{10}(x-50)]    where x\ge 50

Thus, the revenue

R=200x-\frac{1}{2}(x^2-50x)

For Max R

\frac{dR}{dx}=0

\implies 200-\frac{1}{2}\times (2x-50)=0

\implies 200-x+25=0

\implies x=225

Now,

\frac{d^2R}{dx^2}=-1&lt;0

Hence at x=225 Revenue R is maximum

Therefore, number of passengers that will maximize the revenue of the operator is 225

Hope this answer is helpful.

Know More:

Q: Currently, the demand equation for baseball bats is Q = 100 - 2P. The current price is $15 per bat. Is this the best price to charge in order to maximize revenues?

Click Here: https://brainly.in/question/6911540

Similar questions