Question 1120665: An airline flies between 2 major cities daily. It normally transports 81 people a day at a cost of $210 per person. The airline estimates that it will lose 2 passengers for every $6 that it increases the fare. What should the airline charge to gain the maximum income?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It would be easier to use 1 passenger for every $3 raise.
x=number of passengers changing
(81-x)(210+3x)=17010+243x-210x-3x^2
This is -3x^2+33x+17010
x-value for maximum is at x=-b/2a or -33/6 or -5.50
This makes x 75.5, with a fare of $226.50, but we need integer number of passengers. Because this is not linear, we need to look at both sides of the decimal, at 75 and 76.
76 passengers is the optimal number at a fare of $225 each.
81---210
80---213
79---216
78---219
77---222=17094
76---225=17100
75---224=16800
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