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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?

 Apr 8, 2021
 #1
avatar+212 
+1

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

 Apr 8, 2021
 #2
avatar+31826 
+1

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

 Apr 8, 2021

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