# Having a lot of trouble understanding optimization

• Apr 11th 2009, 03:48 PM
Mohdoo
Having a lot of trouble understanding optimization
Hey, all. I'm new here. I have been struggling very much to get this problem figured out, and I am just completely stuck.

I will type it out:

A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $12, average attendance at a game has been 11,000. A market survey indicates that for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the team set the ticket price to maximize revenue from ticket sales? • Apr 11th 2009, 03:57 PM nzmathman Let R be the revenue and x be the decrease in price of each ticket) Then$\displaystyle R = (12-x)(11000+1000x) = 132000 + 1000x - 1000x^2\displaystyle \frac{dR}{dx} = 1000-2000x$Solve for x to make the equation 0 and you will have the value of x that gives maximum revenue. Then the cost of each ticket is$(12-x)
• Apr 11th 2009, 04:00 PM
Mohdoo
Quote:

Originally Posted by nzmathman
Let R be the revenue and x be the decrease in price of each ticket)

Then $\displaystyle R = (12-x)(11000+1000x) = 132000 + 1000x - 1000x^2$

$\displaystyle \frac{dR}{dx} = 1000-2000x$

Solve for x to make the equation 0 and you will have the value of x that gives maximum revenue. Then the cost of each ticket is \$(12-x)

Wonderful. Thank you so much for this! Yet even so, I get the feeling that there is some sort of formula or concept that goes behind this. It seems very structured and well thought out. Is there some sort of concept or formula that goes behind a problem like this?
• Apr 11th 2009, 04:06 PM
nzmathman
Well for a problem like this you need to think and find a formula for what you are being asked to maximise or minimise. In this case (12-x) is the decreased price of each ticket, and (11000+1000x) is the average number of spectators for a particular value of x. So the revenue must equal these two numbers multiplied...

For maxima/minima one always differentiates the formula and equates to 0. If you are familiar with parabolas you will recognise that our formula in this question is an upside down parabola and has a maximum value when the derivative is 0. In other instances a formula has maxima and minima so you need to do extra work to find which is which (Wink)