SOLUTION: SOLVE FULLY. DEFINE VARIABLES, CREATE QUADRATIC EQUATION AND ANSWER PROBLEM. Tickets to a dance cost $4 and the projected attendance is 300 people. For every $0.10 increase in tic

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: SOLVE FULLY. DEFINE VARIABLES, CREATE QUADRATIC EQUATION AND ANSWER PROBLEM. Tickets to a dance cost $4 and the projected attendance is 300 people. For every $0.10 increase in tic      Log On


   

Question 933553: SOLVE FULLY. DEFINE VARIABLES, CREATE QUADRATIC EQUATION AND ANSWER PROBLEM.
Tickets to a dance cost $4 and the projected attendance is 300 people. For every $0.10 increase in ticket price, the school dance committee predicts attendance will decrease by 5 people. Determine price of ticket that produces greatest revenue.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Tickets to a dance cost $4 and the projected attendance is 300 people.
For every $0.10 increase in ticket price, the school dance committee predicts attendance will decrease by 5 people.
Determine price of ticket that produces greatest revenue.
:
Let x = the no. of 10 cent increases
Also
Let x = no. of 5 people reductions with each price increase
:
y = revenue for ticket sales
y = (4+.10x)(300-5x)
FOIL
y = 1200 - 20x + 30x - .5x^2
A quadratic equation
y = -.5x^2 + 10x + 1200
greatest y (revenue) occurs at the axis of symmetry, x = -b/(2a)
x = %28-10%29%2F%282%2A-.5%29
x = +10 price increases for max revenue
therefore the price will be:
4 + 10(.10) = $5 is the price for max revenue
:
No. of tickets then (300 - 10(5)) = 250 people
:
Max revenue: 250 * 5 = $1250