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.
Answers
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operator charge rs 200 per passenger for 50 passenger
Revenue = 200 * 50 = Rs 10000
discount of rs 5 for each 10 passenger in excess of 50
Let say 10P Passenger in excess of 50 then
Charges 200 - 5P
Revenue = (200 - 5P)(50 + 10P)
=> R = 10000 + 2000P - 250P - 50P²
=> R = 10000 + 1750P - 50P²
dR/dP = 1750 - 100P
=> 1750 - 100P = 0
=> P = 17.5
d²R/dP² = - 100 < 0
Hence Maximum revenue at P = 17.5
but P has to be integer so lets Check at
P = 17 & P = 18
Revenue = (200 - 5P)(50 + 10P)
P = 17 => R = (200 - 85) (50 + 170) = 25300
P = 18 => R = (200 - 90) (50 + 180) = 25300
Passengers = 220 or 230 will maximize the revenue of the operator
if We consider 17.5 also
Then R = (200 - 87.5) (50 + 175) = 25312.5
Passengers = 225
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Answer:
10000
Step-by-step explanation:
100÷50×200 =10,000