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## IsTim 3 years ago For an outdoor concert, a ticket price of \$30 typically attracts 5000 people. For each \$1 increase in the ticket price, 100 fewer people will attend. The revenue, R, is the product of the # of people attending and the price per ticket. If the ticket price that maximizes the revenue is \$40, will that change if the concert area holds a max. of 1200 people? Explain.

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1. IsTim

Formula for revenue is R=(5000-100x)(30+x), or R=-100x^2+2000+150 000. Derivative is -200x+2000. 2nd derivative is R"=-200. Max revenue is \$160 000. Ticket price to Max. revenue is \$40.

2. apoorvk

Why that second derivative? You just need to find the the 'x' for the maxima of the revenue function, which is basically equating the first derivative to zero.

3. IsTim

Just giving as much information as possible.

4. apoorvk

and remember 'x' is the *change* in price of each ticket, you still have to add/subtract it to the \$30 to get the actual price of the ticket.

5. IsTim

x=10.

6. apoorvk

Yes! So ticket prices for optimum revenue for a typical sale of about 5000 tickets per day is 40 bucks. Now, if the constraint is that, the area itself can't hold more than 1200 people, then I am losing about 3800 people. so I can basically raise the price of each ticket by 38 bucks (100 people for every buck). So, find the out the revenue now, when x=38.

7. IsTim

How do I know that I am losing about 3800 people?

8. apoorvk

Well, the typical attendance (in the first line of the question) is about 5000. Now when the capacity is reduced to 1200 people, then you lose 5000-1200=3800 people.

9. IsTim

Oh. That logic surpasses me. Give me some time to try to work this out. Thanks for the help.

10. apoorvk

No worries, take it easy ;)

11. IsTim

Wait, how do I show this algebraically?

12. apoorvk

show algebraically what?

13. IsTim

I just got back to this question after goofing around, and I understand up to the 3 800 people part.

14. apoorvk

very well, your goofing around helped you i see. :p yes so, now 38000 aren't there. so for that amount of reduce in the audience, by how much can i afford to raise the ticket prices? it's written that for every 1 buck increase in the ticket price, i lose 1000 people. so if already lost 3800 people, how much can i raise the ticket price by? (because 1200 people are going to attend the play anyway)

15. apoorvk

actually anyway means that till i have an acceptable increase in tkt prices.

16. IsTim

I'm not really sure. So I just tried subbing 3800 as My quantity and sub it into Q=5000-100x, and then solve for x. Then I sub that into x for my price, and solve, were I get \$68 instead.

17. apoorvk

nah!!!! 'x' is the decrease/increase in price per ticket. if i lose 1000 people for every dollar i increase, and i already lost 3800 people, then i can afford to increase the price by 38 dollars per ticket. now profit= (30+x)(5000-100x) 'x' here is 38. find out the max profit!

18. apoorvk

(which occurs for x=38)

19. IsTim

How did I figure out x?

20. apoorvk

i found out 'x' for you and showed you!! okay do you agree, that the '100x' in that equation is the decrease in the no. of attendees?

21. apoorvk

i mean attenders**

22. IsTim

Ok...

23. apoorvk

so get this?

24. apoorvk

100x=3800 = people lost due to the limited capacity of the audience. so, x = 3800/100 = 38!!

25. IsTim

Still wondering how you got =38...

26. IsTim

Where that equation come from? Did we change 5000 to 3800?

27. apoorvk

no!!!! 1200 is th limiting capacity of the auditorium, remember? so, 5000-1200 = 3800 people who we lost in the audience. we can'rt sell tickets to them, so we make up by increasing the prices to such a point that all tickets are sold, as well we can maximise our profit.

28. IsTim

Maybe we should start from the top again.

29. apoorvk

Sure!! Your job to read it all again. -.-

30. apoorvk

so, scroll up.

31. IsTim

That's a rather rude way to put it, but I'll re-read it.

32. apoorvk

Lol, didn't mean to be rude, sorry.

33. IsTim

It's alright. I'm just still confused on what exactly my next step should be.

34. IsTim

Cause I just stated my work a bit, and then we affirmed how the value 3800 was obtained. And then 100x=3800 popped up.

35. apoorvk

"For each \$1 increase in the ticket price, 100 fewer people will attend." so, when 3800 fewer people are attending, what can the increase in the ticket price be?

36. IsTim

Am...I subbing in 3800 as the 'quantity' then?

37. apoorvk

no! in the original equation for profit, Profit = (30+x)(5000-100x) here 100x=3800 and x=38 so put in x=38 and get the max profit for this case.

38. IsTim

Like into the original equation right?

39. apoorvk

yes!

40. IsTim

And from that, my final answer right?

41. IsTim

AFK.

42. apoorvk

yes! what is the answer that you're getting?

43. IsTim

Answer for Price is \$68. Revenue is \$81 600

44. apoorvk

exactly! 30 + the increase, i.e. 38 is \$68. substituting 38 in the original equation for x, gives you \$81,600, which is correct as well.

45. IsTim

Ok. Thank you very much.

46. apoorvk

no worries!

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