1. ## optimization problem

the number of bus riders per day is represented by 1200(1.5-x), where x is the fare in dollars. what fare will maximize total revenue?
I don't really know what to do, because thats all the information i was given.

i know the fare cannot equal 1.50 and that it cannot be greater than 1.50, but it has to be low enough to make the most profit

2. Originally Posted by surffan
the number of bus riders per day is represented by 1200(1.5-x), where x is the fare in dollars. what fare will maximize total revenue?
I don't really know what to do, because thats all the information i was given.

i know the fare cannot equal 1.50 and that it cannot be greater than 1.50, but it has to be low enough to make the most profit

Revenue equals price times quantity.

In your case, price is x and quantity is 1200(1.5-x).

Multiply the two, and optimize the product.

Good luck!

3. After multiplying the two it will, of course, be a quadratic and you can optimize it by completing the square.

4. does that mean that i convert it to 1800-1200x, is there anything else i should be considering?