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?
Hey there, Guest!
So, let's get started:\(10(20+n)\cdot 800(40-n) = 8000(20+n)(40-n) = 8000(800+40n-n^2).\)
Then:
\(n=-(\frac{40}{2\cdot(-1)})=20\)
Therefore, they will charge \(200+20\cdot 10 = 400\) dollars per ticket.
Hope this helped! :)
( ゚д゚)つ Bye
Revenue = 200 * 40 000 = 8 000 000 presently
x = number of 10 dollar increases from 200 dollars
Revenue = (200+ 10x)(40 000 - 400x)
= 8 000 000 - 80 000x + 400 000x - 4000x^2
= -4000 x^2 +320 000x + 8 000 000
Divie thru by 4000 to make it easier to work with
- x^2 + 80x +2000 max will occur at x = - b/2a = - 80/(-2) = 40
or 40 ten dollar increases would make tix 600 dollars