1. **State the problem:** Claudio's revenue from ticket sales is modeled by the quadratic equation $$R = 15000 + 200x - 10x^2$$ where $x$ is the number of $1 increases in ticket price.$ 2. **Identify what is asked:** (a) Initial revenue (when $x=0$). (b) Revenue when $x=3.5$. (c) Maximum revenue. (d) Number of tickets sold and ticket price at maximum revenue. 3. **Recall the formula and rules:** - The revenue function is quadratic and opens downward (since coefficient of $x^2$ is negative). - The vertex of a parabola $ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. - The maximum revenue is the value of $R$ at the vertex. 4. **Calculate initial revenue:** Substitute $x=0$: $$R = 15000 + 200(0) - 10(0)^2 = 15000$$ 5. **Calculate revenue at $x=3.5$:** $$R = 15000 + 200(3.5) - 10(3.5)^2$$ $$= 15000 + 700 - 10(12.25)$$ $$= 15000 + 700 - 122.5 = 15577.5$$ 6. **Find $x$ for maximum revenue:** Given $a = -10$, $b = 200$, $$x = -\frac{b}{2a} = -\frac{200}{2(-10)} = -\frac{200}{-20} = 10$$ 7. **Calculate maximum revenue:** Substitute $x=10$: $$R = 15000 + 200(10) - 10(10)^2$$ $$= 15000 + 2000 - 1000 = 16000$$ 8. **Find ticket price and tickets sold at maximum revenue:** - Ticket price = initial price + $x$ increases = $30 + 10 = 40$ - Tickets sold = initial tickets - 10 tickets per increase * $x$ = $500 - 10(10) = 400$ **Final answers:** - (a) Initial revenue = $15000$ - (b) Revenue at 3.5 increases = $15577.5$ - (c) Maximum revenue = $16000$ - (d) Tickets sold = 400, Ticket price = 40