Question

The hospital sells raffle tickets to raise funds for new medical equipment. Last year, 2000 tickets sold for $24 each. The fund-raising coordinator estimates that for every $1 decrease in price, 125 more tickets will be sold. a) What decrease in price will maximize the revenue? b) What is the price of a ticket that will maximize the revenue? c) What is the maximum revenue? PLEASE SHOW WORK, Thank you!

Answer 1

Answer:

a. $4.00

b. $20.00

c. $ 5000

Step-by-step explanation:

a. The given information are;

The number of tickets sold = 2000

The price of tickets sold = $24

The number extra tickets sold per decrease in price = 125

Let the number of tickets sold = y

The price of each ticket = x

Therefore, given that the fund-raising coordinator estimates that the number of tickets sold vary proportionately with the price of the ticket, we have a linear relationship of the form;

y = mx + c

Where:

m = The slope which is change in y divided by change in x

Change in y = 125

Change in x = -1

m = 125/(-1) = -125

Where the change in y = y - 2000 and change in x = x - 24

we have for a linear straight line relation;

(y - 2000)/(x - 24) = m

Which gives

y - 2000 = -125(x - 24)

y - 2000 = -125·x + 3000

y  = -125·x + 3000 + 2000 = -125·x + 5000

The revenue = Price × Quantity sold = x × y

Where y = -125·x + 5000, we have

Revenue = x×(-125·x + 5000) = -125·x² + 5000·x

The maximum revenue occur at the point where the slope of the revenue = 0, which is d(yx)/dx =d(-125·x² + 5000·x)/dx = -250·x +5000 = 0

x = -5000/(-250) = 20

Therefore, the price that gives maximum revenue = $20 which is a decrease of $24 - $20 = $4.00

b. The price that will maximize the revenue = $20

c. The quantity sold at maximum revenue = -125 × 20 + 5000 = 2500

The maximum revenue = 2500 × 20 = $5000.00