SOLUTION: When tickets at a dance cost $4, the expected attendance is 300 people. for every $0.10 increase in ticket price, the dance committee projects that the attendance will decrease by

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Question 1181585: When tickets at a dance cost $4, the expected attendance is 300 people. for every $0.10 increase in ticket price, the dance committee projects that the attendance will decrease by 5 people. what would be the maximum revenue? what ticket price would maximize the revenue?
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
Ticket price as is = $4
Attendence with ticket price as is = 300
Revenue with ticket price as is = (300)($4) = $1200

Let x = the number of ten-cent increases made in the ticket price.
Then the ticket price will increase by $0.10x, making the 
new higher ticket price be $4+$0.10x.

That will also cause the attendance to decrease by 5x.
Then the new lower attendance will be 300-5x.

Let y = the new higher (hopefully) revenue (the amount of money taken in
from selling all the tickets at the new higher ticket price and the new
lower attendance).



Using FOIL on the right side,









The maximum new attendance y will occur at the vertex of the parabola,
since the parabola opens downward (since the leading coefficient, a, is
negative).  We use the vertex formula


 
So we should make 10 ten-cent increases in the ticket price,
which makes the new higher ticket price $5 and the new lower attendance
50 fewer or only 250. So the new higher revenue will be (250)($5) or $1250.

We could also have substituted 10 for x in



and gotten the same answer $1250.

Edwin


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