<PRE><B>Question 933553</B>
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 = {{{(-10)/(2*-.5)}}}
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
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