To solve part (b), we need to maximize the revenue $R(x)$, which is the product of the price $p(x)$ and the number of spectators $x$. The revenue function is given by:

$R(x) = p(x) \cdot x$

From part (a), we know that the demand function $p(x)$ is:

$p(x) = -\frac{3}{4000}x + 28.75$

So, substituting $p(x)$ into $R(x)$:

$R(x) = \left(-\frac{3}{4000}x + 28.75\right) \cdot x$

Expanding this, we get:

$R(x) = -\frac{3}{4000}x^2 + 28.75x$

To maximize the revenue, we need to find the vertex of this quadratic function since it opens downwards (the coefficient of $x^2$ is negative). The $x$-coordinate of the vertex for a quadratic function $ax^2 + bx + c$ is given by:

$x = -\frac{b}{2a}$

Here:

• $a = -\frac{3}{4000}$
• $b = 28.75$

So,

$x = -\frac{28.75}{2 \cdot \left(-\frac{3}{4000}\right)}$

Calculating this:

$x = \frac{28.75 \times 4000}{6} = \frac{115000}{6} \approx 19166.67$

Now, substitute $x = 19166.67$ back into $p(x)$ to find the ticket price that maximizes revenue:

$p(19166.67) = -\frac{3}{4000} \cdot 19166.67 + 28.75$

Calculating:

$p(19166.67) \approx 14.375$

So, the ticket price should be set at approximately $14.38 to maximize revenue.

Would you like a more detailed breakdown of any step, or have additional questions?

Related Questions

1. What would the revenue be at the optimal ticket price?
2. How would the attendance change if the ticket price were set to maximize revenue?
3. What is the elasticity of demand at the optimal ticket price?
4. How sensitive is the revenue to small changes in ticket price near the optimal level?
5. How would increasing the stadium capacity affect the optimal ticket price?

Tip:

For any function maximization, finding the derivative and setting it to zero can identify critical points, which is especially useful in revenue and cost functions.