1. **State the problem:** Claudio sells 500 tickets at 30 each. For every 1 increase in price, 10 fewer tickets are sold. The revenue function is given by $$R = 15000 + 200x - 10x^2$$ where $x$ is the number of $1$ increases in price. 2. **Identify what is asked:** (a) Initial revenue (b) Revenue at 3.5 price increases (c) Maximum revenue (d) Number of tickets sold and ticket price at maximum revenue 3. **Recall the revenue formula:** Revenue $R = ext{price per ticket} \times \text{number of tickets sold}$. 4. **(a) Initial revenue:** Initial revenue is when $x=0$ (no price increase): $$R(0) = 15000 + 200 \times 0 - 10 \times 0^2 = 15000$$ 5. **(b) Revenue at 3.5 increases:** Calculate $R(3.5)$: $$R(3.5) = 15000 + 200 \times 3.5 - 10 \times (3.5)^2$$ $$= 15000 + 700 - 10 \times 12.25$$ $$= 15000 + 700 - 122.5 = 15577.5$$ 6. **(c) Maximum revenue:** The revenue function is quadratic: $$R = -10x^2 + 200x + 15000$$ (rearranged in standard form). Since the coefficient of $x^2$ is negative, the parabola opens downward, so the vertex is a maximum. The vertex $x$-value is given by: $$x = -\frac{b}{2a} = -\frac{200}{2 \times (-10)} = -\frac{200}{-20} = 10$$ Calculate maximum revenue: $$R(10) = 15000 + 200 \times 10 - 10 \times 10^2$$ $$= 15000 + 2000 - 1000 = 16000$$ 7. **(d) Tickets sold and ticket price at maximum revenue:** Price per ticket at max revenue: $$\text{Price} = 30 + x = 30 + 10 = 40$$ Tickets sold at max revenue: $$\text{Tickets} = 500 - 10x = 500 - 10 \times 10 = 400$$ **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