An airline finds that if it prices a cross-country ticket at $300, it will sell 200 tickets per day. It estimates that each $10 price reduction will result in 50 more tickets sold per day. Find the ticket price (and the number of tickets sold) that will maximize the airline's revenue

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