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
