Question 114385: Tickets to a school dance cost $4, and the projected attendance is 300 persons. It is further projected that for every $0.1 increase in ticket price, the average attendance will decrease by 5. At what ticket price will the receipts from the dance be $1218?
If there is more then one solution, separate your answers with commas .
So I've got the formula of:
(4+.10x)(300-5x)=1218
.50x2+10x-18=0
Now I'm stuck. Can someone please help?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Tickets to a school dance cost $4, and the projected attendance is 300 persons. It is further projected that for every $0.1 increase in ticket price, the average attendance will decrease by 5. At what ticket price will the receipts from the dance be $1218?
If there is more then one solution, separate your answers with commas .
:
So I've got the formula of:
(4+.10x)(300-5x) = 1218
:
You got all the thinking part done ok
:
.50x2+10x-18=0
however, here you should have:
-.5x^2 + 10x - 18 = 0
:
It's easier to deal with if the coefficient of x^2 is +1;
Multiply equation by -2 and you have:
+x^2 - 20x + 36 = 0
:
This factors easily to:
(x-18)(x-2) = 0
x = +18, +2 as the increased ticket prices that has a revenue of $1218
:
:
When x = 18; Ticket price: 4 + 1.80 = $5.80
When x = 2: Ticket price: 4 + .2 = $4.20
:
We can confirm that using your original equation
x = 18,
(4 + 1.8) (300 - 90))
5.8 * 210 = 1218
:
Price of $2
(4 + .2)*(300 - 10)
4.2 * 290 = 1218 also
:
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