Question 468137: A carnival usually sells three thousand 60¢ ride tickets on a Saturday. For each 20¢ increase in price, management estimates that 70 fewer tickets will be sold. What is the smallest increase in ticket price that will produce $2,344.00 of revenue on Saturday
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! they are currently selling 3000 tickets at 60 cents apiece to generate a revenue of 3000 * .6 = $1,800.
if they increase the price to 80 cents a ticket, they will sell 2930 tickets to generate a revenue of 2930 * .8 = $2344.
looks like they can do it with the 20 cents increase in price.
the equation to find the relationship between the raise in price and the drop in the number of tickets was a little tricky to find, but it turns out to be:
R = (3000 - 70*x) * (.6 + .2*x)
R equals the revenue.
x equals the number of times that the number of tickets sold goes down by 70 and also equals the number of times the price goes up by .2.
simplifying that equation gets you:
R = -14x^2 + 558x + 1800
this is a quadratic equation that will rise to a certain point and then start going back down after that.
a graph of this equation looks like the following:
the y value of the graph is the revenue generated.
the x value of the graph is the number of times that the number of tickets goes down by 70 and the price of each ticket goes up by 20 cents (.2).
a closer look at x = {0,1,2} shows that y = 1800 when x = 0, and y = 2344 when x = 1, and y = 2860 when x = 2.
a graph of the closer look is shown below:
the lowest horizontal line is y = 1800.
the next lowest horizontal line is y = 2344.
the next lowest horizontal line is y = 2860.
drop a vertical from each intersection, and you'll see the value of x = 0, 1, and 2 for y values of 1800, 2344, and 2860.
when x = 0, the number of tickets sold was 3000 and the price was 60 cents.
when x = 1, the number of tickets sold was 2930 and the price was 80 cents.
when x = 2, the number of ticket sold was 2860 and the price was 1 dollar.
we reached the revenue requirement when x = 1.
in fact, we reached it exactly when x = 1.
revenue = $2344 when the price goes up by 20 cents and the number of tickets sold goes down by 70.
when x = 2, the price went up another 20 cents to equal $1.00, and the number of tickets sold went down another 70 to equal 2860.
the numbers are:
3000 * .6 = 1800
2930 * .8 = 2344
2860 * 1.0 = 2860
from the first graph, you can see that the revenue will rise until x gets to around 19 or 20.
at that point, the price of the ticket will be .6 + 19*.2 and .6 + 20*.2.
this equates to a price of 4.4 and 4.6.
the number of tickets sold will have dropped down to 3000 - 19*70 and 3000 - 20*70.
this equates to a number of tickets sold at 1670 and 1600.
1670 * 4.4 = 7348
1600 * 4.6 = 7360
1530 * 4.8 = 7344
looks like the peak was at x = 20.
the max/min point of a quadratic equation is given by x = -b/2a
our equation became:
y = -14x^2 + 558x + 1800.
in this quadratic equation:
a = -14
b = 558
c = 1800
the max/min point is at x = -558/-28.
this comes out to be x = 19.92857143
that's the theoretical max/min point.
when x = 19.92857143, y becomes 7360.071429
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