let the number of $10 increases be n
Now:
cost per ticket = 200
number sold = 40,000
after increase:
cost per ticket = 200 + 10n
number sold = 40000 - 400n
revenue = (200+10n)(40000-400n)
= 10(400)(20+n)(100 - n)
= 4000(2000 + 80n - n^2)
d(revenue)/dn = 4000(80 - 2n) = 0 for a max of revenue
n = 40
so they should increase the ticket by 10(40) = 400
new cost of ticket = 200 + 10n = 200+400 = $600
If you don't know Calculus, which I used, find the vertex of
2000 + 80n - n^2
An Ariline currently charges 200 dollars per ticket and sells 40,000 tickets a week. For every 10 dollars they increase the ticket price, they sell 400 fewer tickets a week. How many dollars should they charge to maximize their total revenue?
1 answer
1 answer