A baseball team plays in a stadium that holds 56000 spectators. With the ticket price at $10 the average attendance has been 22000 . When the price dropped to $8, the average attendance rose to 28000 . Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

Step 1 ：Let's denote the ticket price as \(p\) and the number of spectators as \(s\). According to the problem, we know that the relationship between \(p\) and \(s\) is linear, so we can express it as \(s = ap + b\).

Step 2 ：We have two points on this line: \((10, 22000)\) and \((8, 28000)\). We can use these two points to solve for \(a\) and \(b\).

Step 3 ：First, we can set up the system of equations: \[\begin{cases} 22000 = 10a + b\\ 28000 = 8a + b\end{cases}\]

Step 4 ：Solving this system, we get \(a = -3000\) and \(b = 52000\). So the relationship between \(p\) and \(s\) is \(s = -3000p + 52000\).

Step 5 ：The revenue \(R\) is the product of the price and the number of spectators, so \(R = ps = p(-3000p + 52000) = -3000p^2 + 52000p\).

Step 6 ：To find the price that maximizes revenue, we need to find the maximum of this quadratic function. The maximum occurs at the vertex of the parabola, which is at \(p = -\frac{b}{2a} = -\frac{52000}{2(-3000)} = \frac{52}{6} = \frac{26}{3}\approx 8.67\).

Step 7 ：So the ticket price that would maximize revenue is \(\boxed{\$8.67}\).