The manage of a bike shop has found that, at a price ( in dollars ) of p(x)=150-x/4 per byke, x bikes will be sold.
A) Find an expression for the total revenue from the sale of x bike. ( Revenue = demand X price )
B) Find the number of bicycles sales that leads to maximum revenue.
C) Find the maximum revenue.

( Please explain the concepts and every thing because i have a really hard time understand the process. ) Thank you very much!

### One Response to Math/ Business question!?

1. eborg says:

So first lets make a quick correction Revenue ≠ Demand * Price, Revenue = Price * Quantity sold
This leads us to solve for A

A) Revenue = Price * Quantity => Revenue = p(x)*x = 150x - (x^2)/4 = r(x)

For the next question we find a maximization problem for our r(x). If you graph it you'll find a parabola, and you'll be able to see visually what the maximum revenue you can get is (hint: its the top of the hill that gets graphed)

To find the max for this there are a ton of different methods, the easiest is to just take the derivative and set it to zero

so you get r' = 0 = 150 - 2x/4 => x/2 = 150 => x = 300

So you now know B) 300 is the max number of bikes

so for the next one we just know that it's whatever revenue we made when we sold 300 bikes or
C) r(300) = 22,500