SOLUTION: The revenue of a charter bus company depends on the number of unsold seats. IF 100 seats are sold, the price is $50 per seat. Each unsold seat increases the price per seat by $1.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The revenue of a charter bus company depends on the number of unsold seats. IF 100 seats are sold, the price is $50 per seat. Each unsold seat increases the price per seat by $1.      Log On


   

Question 815773: The revenue of a charter bus company depends on the number of unsold seats. IF 100 seats are sold, the price is $50 per seat. Each unsold seat increases the price per seat by $1. Let x represent the number of unsold seats.
(a) write an expression for the number of seats sold.
(b) Write an expression for the price per seat.
(c) Write an expression for the revenue.
(d) Find the number of unsold seats that will produce the maximum revenue.
(e) Find the maximum revenue.
I've come up with the expression for (b) as follows p=50+x, but I'm not sure if that is correct. We are working on applications of quadratic functions and I don't understand the conecpts very well at all.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

IF 100 seats are sold, the price is $50 per seat.

Each unsold seat increases the price per seat by $1. 

Let x represent the number of unsold seats. 

(a) write an expression for the number of seats sold. (100-x)

(b) Write an expression for the price per seat. %2850%2B+1%28x%29%29

(c) Write an expression for the revenue. R = (100-x)(50+x)

(d) Find the number of unsold seats that will produce the maximum revenue.

    Revenue = 5000 + 50x - x^2  = -(x-25)^2 +625+ 5000 = -(x-25)^2 + 5625

Revenue= -(x-25)^2 + 5625 

Parabola V(25,5625)opening downward

  25 unsold tickets produces the maximum revenue

(e) Find the maximum revenue. $5625