THE QUESTION IS IN TUE DESCRIPTION! HELP!

Question

You are a business owner. You would like to know how many tickets you need to sell for a show to make a profit. Currently you are selling each ticket for $5.00, but for every $0.50 increase in the price, 100 fewer people will attend. Your club/auditorium has 1200 seats, what ticket price will maximize your profit to be successful. You will need to set up and solve the equation properly, then graph your maximized profit.

Answer 1

You probably are looking at maximizing the revenue, not profit

 

But here we go

 

Assuming all seats will be sold at the $5 price you would have a revenue of $6000

 

 

Revenue = (1200 - 100X) (5+.5X)

 

1200- 100X    how many will attend for each 50 cent increase, X = 1 means 1 , 50 cent increase

5+.5X    what each person will pay that attends

 

R(X) = (1200 - 100X)(5+ .5X)  = 6000 + 100X - 50X^2

 

This is a quadratic opening down and has a maximum at it's vertex:

 

-b/2a

 

-100/(-100) = 1

 

The max occurs at X = 1, 1 50 cent increase

 

Therefore the ticket price they should charge should be $5.50

 

 

To graph the quadratic, you need to graph in the first quadrant with a y-int of (0,6000).
Then just plug in values of X into the quadratic above and plot points.
Make sure to plot (1, 6050) as this is the Maximum
When X = 12, no one is attending so make sure to plot that as your x-int.