An airline flying to a Midwest destination can sell 20 coach-class tickets per day at a price of $250 and ten business-class tickets per day at a price of $750. It finds that for each $10 decrease in the price of the coach ticket, it will sell four more per day, and for each $50 decrease in the business-class price, it will sell two more per day. What prices should the airline charge for the coach and business-class tickets to maximize revenue? [Hint: Let x be the number of $10 price decreases for coach tickets and y be the number of $50 price decreases for business-class tickets, and use the methods in this example and this example for each type of car.]

Solution :

Let x = number of coach class tickets sold per day on $ 10 reduced price

y = number of the business class tickets sold per day on $ 50 reduced per price.

Number of the coach class tickets sold = 20 per day at a price of $ 250

Number of the business class tickets sold = 10 per day at a price of $ 750

Decrease in the price of coach class tickets by $ 10 increases the number of tickets sold by 4 per day

Decrease in the price of business class tickets by $ 50 increases the number of tickets sold by 2 per day.

Therefore, the coach price, p(x) = 250 - 10x

Business class price, p(y) = 750 - 50y

Coach tickets, q(x) = 2 + 4x

Business class tickets, q(y) = 6 + 2y  

Revenue R(x, y) = p(x) q(x) + q(y) q(y)

R(x, y) = (250-10x)(20+4x)+(750-50y)(6+2y)

R(x) = (250-10x)(0.4)+(20+4x)(-10)+0 = 0

1000-40x-200-40x=0

80x = 800

x = 10

R(y) = 0 + (750 - 50y)(0.2) + (6+2y)(-50) = 0

1500 - 100y+(-300-100y) = 0

200 y = 1200

y = 6

P(x) = 250 - 10(10) = $ 150

P(y) = 780-50(6) = $450

q(x) = 20 + 4(10) = 60

q(y) = 6 + 2(6) = 18

Maximum revenue = 150 x 60 + 450 x 18

                              = 9000 + 8100

                              = $ 17100

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation: