SOLUTION: Tickets to a school dance cost $4 and the projected attendance is 300 people. For every $1 increase in the ticket price, the dance committee projects that attendance will decrease

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Question 1153135: Tickets to a school dance cost $4 and the projected attendance is 300 people. For every $1 increase in the ticket price, the dance committee projects that attendance will decrease by 5 attendees. Determine the dance committee’s greatest possible revenue. What ticket price will produce the
greatest revenue?
Found 3 solutions by ikleyn, josmiceli, ankor@dixie-net.com:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

The formula for the number of attendees as the function of price is this

    n(p) = 300 - 5*(p-4)

where p is the price in dollars  and n(p) is the number of attendees.



    Check it on your own and make sure that you understand this formula !



Then the revenue is the product of price by the number of attendees


    R(p) = p*n(p) = p*(300 - 5*(p-4)) = 300p - 5p^2 + 20p = -5p^2 + 320p.    (1)


So, the revenue (1) is the quadratic function  R(p) = - 5p^2 + 320p.


Every quadratic function f(x) = ax^2 + bx + c  with the negative leading coefficient "a" has the maximum at


    x = ;  in your case  the optimum price is  p =  =  = 32 dollars.


At this price, the number of attendees will be  n(32) = 300 - 5*(32-4) = 160  and the revenue will be 32*160 = 5120 dollars

against the 4*300 = 1200 dollars that the committee has today (!)


Do you see the difference ?

Solved.

-------------------

On similar problems, see the lessons
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola
    - A rectangle with a given perimeter which has the maximal area is a square
    - A farmer planning to fence a rectangular garden to enclose the maximal area
    - A farmer planning to fence a rectangular area along the river to enclose the maximal area
    - A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
    - Finding the maximum area of the window of a special form
    - Using quadratic functions to solve problems on maximizing revenue/profit (*)
    - OVERVIEW of lessons on finding the maximum/minimum of a quadratic function

A convenient place to observe all these lessons from the  "bird flight height"  is the last lesson in the list.

Pay special attention to the lesson marked (*) in the list --- it is the closest lessons by its problems (!)


Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = the number of $1 increases in ticket price
Let = total revenue from the tickets
---------------------




The formula for is



Plug this result back into equation to get



-------------------------
The ticket price that gives this maximum revenue is:

$32 per ticket
-------------------------
check:
Here's the plot:
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 people.
For every $1 increase in the ticket price, the dance committee projects that attendance will decrease by 5 attendees.
Determine the dance committee’s greatest possible revenue. What ticket price will produce the greatest revenue?
:
Let x = no. of $1 increases and no. of 5 less attendees
Revenue = ticket price * no. of attendees
R(x) = (4 + x) * (300 - 5x)
FOIL
R(x) = 1200 - 20x + 300x - 5x^2
a quadratic equation
R(x) = -5x^2 + 280x + 1200
Max rev occurs on the line of symmetry, using x = -b/(2a)
x =
x = 28 ea $1 increases will give max revenue
That would be: 4 + 28 = $32 a ticket
:
No. attending the dance: 300 - (5*28) = 160 attendees
:
Max revenue: 32 * 160 = $5120 vs only $1200 at original price
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