Problem of the Day


A basketball team plays in a stadium that holds 60,000 people. With ticket prices at $28, the average attendance had been 32,000. After ticket prices were lowered to $24, the average attendance rose to 36,000. Assuming that the demand for tickets is a linear function of ticket prices, what price should the team charge for tickets to maximize its revenue?

A. $26
B. $28
C. $30
D. $32
E. $35

Reveal Answer

Answer

C. $30

See the Solution

Solution

[latexpage]

Attendance is a linear function of ticket prices, and we have two points that satisfy that function: (28,32000) and (24,36000). ¬†Using these points, we can calculate the slope of the function to be -1,000 and the y-intercept to be 60,000. So, the number of people who attend, $A(x)$ when the tickets are $x$ dollars a piece, is $A(x)=60,000-1,000x$. ¬†The team’s revenue is equal to their attendance multiplied by their ticket price, so the revenue function is $R(x)=60,000x-1,000x^2$. We need to find the value of $x$ that maximizes this function.

$R(x)=60,000x-1,000x^2$

$R(x)=-1,000(x^2-60)$

Now, we can complete the square as follows:

$R(x)=-1,000(x^2-60+900)+900,000$

$R(x)=-1000(x-30)^2+900,000$

Clearly, this will be maximized when $x=30$.

 


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