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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
+212
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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
+31826
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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