Answer:
Ticket price = $170
Number of tickets sold = 850
Step-by-step explanation:
Let X= the number of $10 price reduction and the number of 50 tickets increase
Let p= price per ticket
Let q= number of tickets sold
Let R= revenue
Let p(x) = price of ticket now sold
Let q(x) = number of tickets now sold
R(x) = p(x) * q(x)
P(x) = 300 - 10x
q(x)= 200 + 50x
R(x) = (300 -10x) (200 - 50x)
= 60000 + 15000x - 2000x - 500x^2
= 60000 + 13000x - 500x^2
Differentiate with respect to x
dR(x)/dx = 13000-1000x
Put dR(x)/dx = 0
0= 13000 - 1000x
1000x = 13000
Divide through by 1000
1000x/1000 = 13000/1000
x = 13
Therefore x = 13 for $10 price reduction and 50 tickets increase
p(x) = 300 - 10x
= 300 - 10*13
= 300 - 130
= $170
q(x) = 200 + 50x
= 200 + 50*13
= 200 + 650
= 650
Maximum Revenue = p(x) * q(x)
= 170 * 850
= $144,500
For 13 reductions, we have $170 price reduction and 850 ticket sales
In order to maximize revenue the ticket price should be $ 170, with which 850 tickets would be sold and $ 144,500 profit would be generated.
Since an airline finds that if it prices a cross-country ticket at $ 300, it will sell 200 tickets per day, and it estimates that each $ 10 price reduction will result in 50 more tickets sold per day, to find the ticket price (and the number of tickets sold) that will maximize the airline's revenue, the following calculations must be performed:
Therefore, in order to maximize revenue the ticket price should be $ 170, with which 850 tickets would be sold and $ 144,500 profit would be generated.
Learn more in https://brainly.com/question/8984066
