IsTim
  • IsTim
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.
Mathematics
katieb
  • katieb
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IsTim
  • 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.
apoorvk
  • 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.
IsTim
  • IsTim
Just giving as much information as possible.

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apoorvk
  • 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.
IsTim
  • IsTim
x=10.
apoorvk
  • 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.
IsTim
  • IsTim
How do I know that I am losing about 3800 people?
apoorvk
  • 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.
IsTim
  • IsTim
Oh. That logic surpasses me. Give me some time to try to work this out. Thanks for the help.
apoorvk
  • apoorvk
No worries, take it easy ;)
IsTim
  • IsTim
Wait, how do I show this algebraically?
apoorvk
  • apoorvk
show algebraically what?
IsTim
  • IsTim
I just got back to this question after goofing around, and I understand up to the 3 800 people part.
apoorvk
  • 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)
apoorvk
  • apoorvk
actually anyway means that till i have an acceptable increase in tkt prices.
IsTim
  • 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.
apoorvk
  • 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!
apoorvk
  • apoorvk
(which occurs for x=38)
IsTim
  • IsTim
How did I figure out x?
apoorvk
  • 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?
apoorvk
  • apoorvk
i mean attenders**
IsTim
  • IsTim
Ok...
apoorvk
  • apoorvk
so get this?
apoorvk
  • apoorvk
100x=3800 = people lost due to the limited capacity of the audience. so, x = 3800/100 = 38!!
IsTim
  • IsTim
Still wondering how you got =38...
IsTim
  • IsTim
Where that equation come from? Did we change 5000 to 3800?
apoorvk
  • 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.
IsTim
  • IsTim
Maybe we should start from the top again.
apoorvk
  • apoorvk
Sure!! Your job to read it all again. -.-
apoorvk
  • apoorvk
so, scroll up.
IsTim
  • IsTim
That's a rather rude way to put it, but I'll re-read it.
apoorvk
  • apoorvk
Lol, didn't mean to be rude, sorry.
IsTim
  • IsTim
It's alright. I'm just still confused on what exactly my next step should be.
IsTim
  • 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.
apoorvk
  • 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?
IsTim
  • IsTim
Am...I subbing in 3800 as the 'quantity' then?
apoorvk
  • 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.
IsTim
  • IsTim
Like into the original equation right?
apoorvk
  • apoorvk
yes!
IsTim
  • IsTim
And from that, my final answer right?
IsTim
  • IsTim
AFK.
apoorvk
  • apoorvk
yes! what is the answer that you're getting?
IsTim
  • IsTim
Answer for Price is $68. Revenue is $81 600
apoorvk
  • 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.
IsTim
  • IsTim
Ok. Thank you very much.
apoorvk
  • apoorvk
no worries!

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