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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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Just giving as much information as possible.

  4. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    x=10.

  6. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    How do I know that I am losing about 3800 people?

  8. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    Oh. That logic surpasses me. Give me some time to try to work this out. Thanks for the help.

  10. apoorvk
    • 3 years ago
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    No worries, take it easy ;)

  11. IsTim
    • 3 years ago
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    Wait, how do I show this algebraically?

  12. apoorvk
    • 3 years ago
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    show algebraically what?

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

  14. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    actually anyway means that till i have an acceptable increase in tkt prices.

  16. IsTim
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    (which occurs for x=38)

  19. IsTim
    • 3 years ago
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    How did I figure out x?

  20. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    i mean attenders**

  22. IsTim
    • 3 years ago
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    Ok...

  23. apoorvk
    • 3 years ago
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    so get this?

  24. apoorvk
    • 3 years ago
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    100x=3800 = people lost due to the limited capacity of the audience. so, x = 3800/100 = 38!!

  25. IsTim
    • 3 years ago
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    Still wondering how you got =38...

  26. IsTim
    • 3 years ago
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    Where that equation come from? Did we change 5000 to 3800?

  27. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    Maybe we should start from the top again.

  29. apoorvk
    • 3 years ago
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    Sure!! Your job to read it all again. -.-

  30. apoorvk
    • 3 years ago
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    so, scroll up.

  31. IsTim
    • 3 years ago
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    That's a rather rude way to put it, but I'll re-read it.

  32. apoorvk
    • 3 years ago
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    Lol, didn't mean to be rude, sorry.

  33. IsTim
    • 3 years ago
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    It's alright. I'm just still confused on what exactly my next step should be.

  34. IsTim
    • 3 years ago
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    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
    • 3 years ago
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    "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
    • 3 years ago
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    Am...I subbing in 3800 as the 'quantity' then?

  37. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    Like into the original equation right?

  39. apoorvk
    • 3 years ago
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    yes!

  40. IsTim
    • 3 years ago
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    And from that, my final answer right?

  41. IsTim
    • 3 years ago
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    AFK.

  42. apoorvk
    • 3 years ago
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    yes! what is the answer that you're getting?

  43. IsTim
    • 3 years ago
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    Answer for Price is $68. Revenue is $81 600

  44. apoorvk
    • 3 years ago
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    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
    • 3 years ago
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    Ok. Thank you very much.

  46. apoorvk
    • 3 years ago
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    no worries!

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