A tour operator charges rupees 200 per passengers with discount of rupees 5 for is 10 passengers in excess of 50. determine the numbers of passengers that will maximize the revenue of the operator
Answers
Answered by
1
Answer:
Let the no of passengers = 50+X
The ticket rate per passenger= 200- (X/10)•5 = 200- X/2
Revenue, R= (50+X) { 200-X/2} = 10000+200X-25X - (X^2/2) = 10000+175X -(X^2/2).
To get the critical point, find dR/dX &equate to Zero.
dR/dX = 175-(2X/2) = 0
X= 175
Now take the derivative AGAIN & check if it is NEGATIVE.
d^2 R/dX^2 = -1 ,which is negative. Hence R a7575ttains maximum when X= 175
Hence, the no of passengers to maximize R is equal to 50+175= 225
The max Revenue = 225×115 = 25875.
Similar questions