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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.
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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IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Formula for revenue is R=(5000100x)(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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why 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
 3 years ago
Best ResponseYou've already chosen the best response.0Just giving as much information as possible.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes! 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
 3 years ago
Best ResponseYou've already chosen the best response.0How do I know that I am losing about 3800 people?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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 50001200=3800 people.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Oh. That logic surpasses me. Give me some time to try to work this out. Thanks for the help.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No worries, take it easy ;)

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Wait, how do I show this algebraically?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0show algebraically what?

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0I just got back to this question after goofing around, and I understand up to the 3 800 people part.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0very 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually anyway means that till i have an acceptable increase in tkt prices.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0I'm not really sure. So I just tried subbing 3800 as My quantity and sub it into Q=5000100x, and then solve for x. Then I sub that into x for my price, and solve, were I get $68 instead.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nah!!!! '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)(5000100x) 'x' here is 38. find out the max profit!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(which occurs for x=38)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0100x=3800 = people lost due to the limited capacity of the audience. so, x = 3800/100 = 38!!

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Still wondering how you got =38...

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Where that equation come from? Did we change 5000 to 3800?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no!!!! 1200 is th limiting capacity of the auditorium, remember? so, 50001200 = 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
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe we should start from the top again.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sure!! Your job to read it all again. .

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0That's a rather rude way to put it, but I'll reread it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lol, didn't mean to be rude, sorry.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0It's alright. I'm just still confused on what exactly my next step should be.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Cause I just stated my work a bit, and then we affirmed how the value 3800 was obtained. And then 100x=3800 popped up.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"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
 3 years ago
Best ResponseYou've already chosen the best response.0Am...I subbing in 3800 as the 'quantity' then?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no! in the original equation for profit, Profit = (30+x)(5000100x) here 100x=3800 and x=38 so put in x=38 and get the max profit for this case.

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Like into the original equation right?

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0And from that, my final answer right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes! what is the answer that you're getting?

IsTim
 3 years ago
Best ResponseYou've already chosen the best response.0Answer for Price is $68. Revenue is $81 600

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exactly! 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.
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