# 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

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). Â 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\$. Â 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\$.