# Thread: optimization and revenue problem

1. ## optimization and revenue problem

A baseball team plays in a stadium that holds 62000 spectators. With the ticket price at $12 the average attendence has been 26000. When the price dropped to$11, the average attendence rose to 31000. Assume that attendence is linearly related to ticket price.

What ticket price would maximize revenue?

2. Originally Posted by eawolbert
A baseball team plays in a stadium that holds 62000 spectators. With the ticket price at $12 the average attendence has been 26000. When the price dropped to$11, the average attendence rose to 31000. Assume that attendence is linearly related to ticket price.

What ticket price would maximize revenue?
we have a linear relationship, and we are given two points.

let the ticket price be plotted on the x-axis and the attendance on the y-axis. we have the points (12,26000) and (11,31000). we can use this to get the equation of the line that relates the ticket price, P, and the attendance, A. This will be your constraint equation (you know what that is, right?)

Now, the revenue, R, is given by:

R = PA ............the objective equation

you want to maximize R. You must find its derivative and set it to zero and solve for the appropriate quantity of whatever variable you use. of course, you use the constraint equation to solve for one variable in terms of the other and then plug in the expression in the objective equation so that you get the equation in one variable. then proceed to integrate

can you continue?

3. ok now i'm totally lost can you explain it to me a little more detailed please. thanks

4. Originally Posted by eawolbert
ok now i'm totally lost can you explain it to me a little more detailed please. thanks
ok, we can do this step by step. start by finding the equation of the line between those two points. do you know how to do this?

5. I tried doing that but got a really weird equation. I got -5000x+86000

6. Originally Posted by eawolbert
I tried doing that but got a really weird equation. I got -5000x+86000
That is correct. so you have

A = -5000P + 86000 ....................................(1)

now the revenue is the cost of each ticket times the number of tickets sold, so,

R = PA .............................................(2)

but we need to maximize this. to do that, we need to differentiate. this is a problem, since we have two variables. But we're in luck. from equation (1), we can write one variable in terms of another. so plug in (1) into (2), we get:

R = P(-5000P + 86000)

=> R = -5000P^2 + 86000P

now this last equation is what you want to maximize. can you continue?

7. Awesome. Great that's the part I was missing. thanks a lot

8. Originally Posted by eawolbert
Awesome. Great that's the part I was missing. thanks a lot
you're welcome, dude(?)!