On a certain route, an airline carries 8000 passengers per month, each paying S6

Question

On a certain route, an airline carries 8000 passengers per month, each paying S60. A market survey indicates that for each SI increase in the ticket price, the airline will lose 100 passengers. Find the ticket price that will maximize the airline's monthly revenue for the route. What is the maximum monthly revenue? Hint: first find the linear demand function p(x) which expresses price per ticket as a function of the number of tickets sold. The ticket price that maximizes the monthly revenue is $| |.The maximum monthly revenue is $| |.

Explanation / Answer

R = (8000 - 100x)(60 + x)

= 480,000 + 2000x - 100x^2

For maximum revenue,

R ' = 0

2000 - 200x = 0

x = 10

Ticket price will be 60 + 10

= $ 70

Revenue = (8000 - 100(10) ) * 70

= 7000 * 70

= $ 490, 000

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