Modeling w/ Quadratic Functions

Question

A baseball team plays in a stadium that holds 59,000 spectators. With the ticket price at $10, the average attendance at recent games has been 24,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.

a) Find a function that models the revenue in terms of ticket price. (Let x represent the price of a ticket and R represent the revenue.)

R(x) = ?

b) Find the price that maximizes revenue from ticket sales.

$ = ?

c) What ticket price is so high that no revenue is generated?

$ = ?

Answer 1

Hi Anna!

Here is how I would approach this problem (final answers are bolded):

Part A)

In this part of the problem, we are asked to set up an equation. Any time this is asked, my first step is to define any relevant variables. I will use the following variables:

X = ticket price in dollars

N = the number of attendees

R = revenue in dollars

Although the variable N is not explicitly defined in the problem, I define it here because it is a parameter that affects our revenue R.

Next, we want to see how these variables relate to each other.

It makes sense that the revenue will be equal to the number of tickets sold multiplied by the price of the tickets. Thus, as a first equation, we can write:

R = NX

This equation defines our revenue R in terms of our other 2 variables N and X. But the answer requires us to find a function that relates R to X and X only. The next step then is to find n in terms of X--or in other words, how the number of attendees changes with ticket price.

We are told that when X = 10, there are 24,000 attendees. We will use this information as our baseline:

When x = 10, N = 24,000

How does this function change as X decreases from 10?

For every dollar that price decreases, attendance decreases by 3000. The amount that the ticket price decreases (from our baseline of $10) is not X, but rather 10 - X . (This expression represents the difference between the our new ticket price and the old $10.) This makes logical sense because if the ticket price is $8, the decrease in ticket price is equal to 10 - 8, or $2.

Thus, the amount the attendees increases is equal to 3000*decrease in ticket price or 3000*(10 - X).

Using this information, we can write the following equation:

N = 24,000 + 3000(10 - x)

Or simplified:

N = 24,000 + 30,000 - 3000x

N = -3,000x + 54,000

We have now found our number of attendees N in terms of X and can plug N back into our revenue function:

R = NX

R = (-3,000X + 54,000) * X

R = -3,000X² + 54,000X

Part B)

This next part involves some Calculus, specifically Optimization. We will examine the Revenue function and its derivative to find maximum revenue.

Because its highest order term is raised to a power of 2, and because the coefficient on this term is negative, we know that, when graphed, our revenue function will look like a parabola that opens downwards. The vertex of the parabola (the highest point) will thus represent that highest amount of revenue possible. At this point, the instantaneous slope (or tangential line) will be equal to zero.

This makes sense from a derivative point of view because the slope (derivative) of the revenue curve has changed from positive to negative, and thus must go through zero. Revenue increases up to this point and decreases after.

To find the price that yields the highest revenue, we must find this point where the derivative of the revenue function is equal to zero. (This is how much of optimization in Calculus works).

We start off by finding the equation for the derivative. This revenue function is fairly straightforward so we will only need to use the power rule.

R = -3,000X² + 54,000X

R' = 2 * (-3,000X) + 54,000

R' = -6,000X + 54,000

Then we set this derivative function equal to zero and solve for X:

0 = -6,000X + 54,000

6,000X = 54,000

X = 9

Maximum revenue occurs when the price of a ticket is $9.

(If you have access to a graphing calculator, you can confirm this answer by graphing the revenue curve and using the "calculate maximum function" (available on most graphing calculator) that will tell you where the maximum revenue occurs and what its value is. If you don't have access, no problem! The calculator basically just repeats what we did by hand.)

Part C)

To find the ticket price that will generate no revenue, we set our revenue function equal to 0:

R = -3,000X² + 54,000X

0 = -3,000X² + 54,000X

0 = x(-3000X + 54000)

So 2 possible solutions:

X = 0

-3000X + 54,000 = 0

-3,000X = -54,000

X = 18

We find that for revenue to be 0, X = 0 or X = 18. In this case, X = 18 is clearly the answer we are looking for: When the ticket price is $18, the price is so high that there will be no revenue.

Hope this explanation helped!