Hospital sells raffle tickets to raise fund for new medical equipment. Last year, 2000 tickets were sold for $24 each. The fund-raising coordinator estimates that for every $1 decrease in price, 125 more tickets will be sold 1) What decreases in price will maximize the revenue? 2) What is the price of a ticket that will maximize the revenue? 3) What is the maximum revenue?
Accepted Solution
A:
Answer:1) Price decrease = $4; 2) new price = $20; 3) maximum revenue = $50 000 Step-by-step explanation:The hospital sold 2000 tickets for $24 each
Revenue = price per ticket × number of tickets sold
Let x = change in price
New price = 24 - x
New number of tickets sold = 2000 + 125x
1) Calculate change in price to maximize revenue
y = (24 - x)(2000 + 125x)
y = 48 000 + 1000x - 125x²
y = -125x² + 1000x + 48 000
a = -125; b = 1000; c = 48 000
The vertex is at
[tex]x = -\dfrac{b}{2a} = -\dfrac{1000}{2(-125)}= \dfrac{1000}{250} = \mathbf{4}[/tex]A price decrease of $4 will maximize revenue.
2) New ticket price
Original price = $24
Price change = - 4
New price = $20
A ticket price of $20 will maximize revenue.
3) Maximum revenue
Ticket price = $20
No of tickets sold = 2000 + 125(4) = 2000 + 500 = 2500
Revenue = 2500 × $20 = $ 50 000
The maximum revenue is $50 000.
The graph below slows the relation between the price drop and total revenue. A price drop of $4 results in a maximum revenue of $50 000.