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# Quadratics A Real World Business Example

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by

## Lynn Ross

on 11 April 2015

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#### Transcript of Quadratics A Real World Business Example

Quadratics:
A Real-World Business Example

By: Lisette M. and Byron S.
Remember:
Quadratics have an x-squared term.
Standard form is:
y = ax^2 + bx + c
When graphed:
Take the shape of a
parabola.
Can be used to model real-world problems where the rate increases (or decreases) until it reaches it maximum (or minimum) point and then decreases (or increases).
Our example will address a business problem related to movie theater ticket prices, attendance, and maximizing revenue.
A movie theater has individual theaters that will hold a combined 1200 customers; however, for the past several months, only about 900 customers come each week.
Currently, the movie theater charges \$11 for general admission. The owners are considering lowering the price to attract more customers.
The owner conducted a survey in the town and the results showed that for every \$1 the ticket price was lowered, the number of local residents who said they would start coming to the movies increased by 100.

Our goal is determine the largest ticket price decrease to achieve the highest attendance and thus the highest revenue amount.
We will let
x = the amount the ticket price is reduced.

The new ticket price would be 11 – x.
(Current price of \$11 minus the price reduction amount.)

Since revenue is found by multiplying the ticket price
with the attendance:
Revenue = (11 – x)(900 + 100x)

Using FOIL to simplify and replacing “Revenue” with y,
we get our function:
y = 9900 + 1100x – 900x – 100x^2
y = -100x^2 + 200x + 9900

Now, since we are looking at the highest or maximum revenue amount, we need the vertex of our parabola.

The formula for finding the vertex is x = (-b)/2a .

In our function, a = -100, b = 200, and c = 9900. Substituting, we get:
x = (-200)/(2(-100)) = (-200)/(-200) =
1

This means that revenue will reach its maximum when the ticket prices are lowered by \$1 to \$10.
The current revenue at \$11 per ticket and an average
attendance of 900 is \$9900.

The new revenue amount, at \$10 per ticket and an attendance that has increased by 100 people to 1000, is \$10,000.

We did some “checks” to make sure our number made sense:

(1) Decrease by \$1 more would make the new ticket price \$9 and the new attendance amount 1100. The revenue would be only \$9900.

(2) Increase the original price by \$1 would make the new ticket price \$12 and the new attendance amount 800 (decrease by 100 instead of increase). The revenue would be only \$9,600.

These checks help us to see that our answer is reasonable.
Here is the graph of the function:
Notice the y-intercept is the "initial attendance" of 900.
In summary, by using a quadratic function to model
the relationship between revenue and a decrease in ticket price, we were able to determine the ticket price reduction that would result in an attendance amount that would generate the highest possible revenue.
Of course, the owners will only lower the price as long the increase in attendance will increase their revenue.
The increase in attendance would be
100x
.

(This is 100 for every dollar, x, that the ticket price is decreased.)
The new attendance amount would be
900 + 100x
.

(The current attendance increased by the additional customers.)
The vertex (1, 10000) shows that by decreasing the ticket price by \$1, the revenue is \$10,000.
Notice that if the ticket price is increase by \$11, the revenue would be \$0. This means that the theater would not sell any tickets at a price of \$22.
We only need quadrant I of the graph where x and y are positive, since negative revenue and negative increase do no make sense in the context of this problem.
Full transcript