Can anyone help me to understand how I would go about solving this max/min problem?

A hall charges $30 per person for a sprts banquet when 120 people attend. For every 10 extra people that attend, the hall will decrease the price by $1.50 per person. What number of people will maximize the revenue for the hall?

I don't know if I understand this correctly but for them to make the most money that would mean so many people would attend that the cost of admission would go to $0? If so then it is simple,

1.50 can go into 30 20 times, 30/1.5=20

This means it will take 20 sets of 10 people to make the admission go to zero.

20 times 10 is 200 people over the 120 people who already paid the 30.

120+200=320

So, if 320 people show up they will make the most money, any more and they get in for free.

Isn't it 0 when 320 is reached? So the max would be 319 x 1.50? But that's only about 480, so what about the first 30 x 120 = 3600? I don't remember the way to solve this with equations, but don't think the above answer is it.
30 x 120 = 3600
21 x 180 = 3780
15 x 220 = 3300 so it's somewhere in there,

It's asking for maximum amount of people not money. The first 120 get in for 30. The next 10 get in for 28.50. The next 10 for 27.00 and so on.

It takes 20 sets of ten to make 30 go to 0.

Therefor 200 people get in for a discount, the last ten people get in for 1.50. The next ten would get in for free.

120+200=320 people

Hm...

For 120 people, the price is 30*120 = 3600

For x people, provided x < 120, the revenue = 30x

But since we know thatwe can get more revenue when there is more than 120 people, the formula that interests us is as follows:

For another x people, provided x > 120, the price of the ticket = 30 - 1.5(\frac{x - 120}{10})

120 - x for the extra people present, and divided by 10 since the price changes for every additional 10 people. Further more, (\frac{x - 120}{10}) must be an integer and is only rounded down.

The revenue then becomes: R = x(30 - 1.5(\frac{x - 120}{10})) = 30x - 1.5x(\frac{x - 120}{10}) = 48x - 0.15x^2

Now, to find the maximum is when the derivative is equal to zero.

\frac{dR}{dx} = 48 - 0.3x

When this equal to zero, x will be:

48 - 0.3x = 0

x = \frac{48}{0.3} = 160

So, when you have 160 people, you get maximum revenue.

The revenue is: R = 48(160) - 0.15(160)^2 = 3840

Well... there is a little trap in this question. Since the price remains the same if there is less than a multiple of 10 more people, the maximum revenue occurs when x = 169, that is R =(169)(30 - 1.5(\frac{169 - 120}{10})) = $4056

One more person, and the revenue becomes $3825, which peaks at $4027.50 for 179 people.

kpg0001, I believe it's like you make reservations for the banquet first, without paying. Depending on the number of persons who are going to attend, the organisers will then adjust the price of the ticket, on the day/night the banquet will take place.

Logically, if there was a case where you describe it, they wouldn't have anybody in the banquet because no one wants to be within the first 120 persons to come (except if the initial price is already very cheap)

If you want to find out how much money that is then:

120x30+

I believe you can take the integral of 10(30-1.5x) from 1-20 to get the right answer

or

you can do it the long way and do (30-1.5)10+(30-3)10 and so on till (30-30)10.

Well the correct answer was posted while I tried another failed attempt. Good job unknown, I learn more than I help ha.

Well, those are the little questions that I'm looking for to keep me sharp ;)

I guess I need to spread some rep around but I was saying I guess I am delusionally(made up word) sharp. Did you look at the question I answered about the quadratic equations?