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