Answer:
a. $4.00
b. $20.00
c. $ 5000
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