1. **Problem statement:** We want to find the optimal ticket price to maximize total revenue and total profit based on attendance changes. 2. **Given data:** - Initial attendance $A_0 = 20000$ when ticket price $p_0 = 100$ - Attendance increases by 2000 for every $5$ decrease in ticket price - Service cost per attendee = $20$ 3. **Define variables:** Let $x$ be the number of $5$ dollar decreases from the original price. 4. **Attendance function:** $$A(x) = 20000 + 2000x$$ 5. **Ticket price function:** $$p(x) = 100 - 5x$$ 6. **Total revenue function:** $$R(x) = A(x) \times p(x) = (20000 + 2000x)(100 - 5x)$$ 7. **Expand revenue:** $$R(x) = 20000 \times 100 - 20000 \times 5x + 2000x \times 100 - 2000x \times 5x$$ $$= 2,000,000 - 100,000x + 200,000x - 10,000x^2$$ $$= 2,000,000 + 100,000x - 10,000x^2$$ 8. **Maximize revenue:** Take derivative and set to zero: $$\frac{dR}{dx} = 100,000 - 20,000x = 0$$ $$20,000x = 100,000$$ $$x = 5$$ 9. **Optimal ticket price for max revenue:** $$p = 100 - 5 \times 5 = 75$$ 10. **Total profit function:** Profit = Revenue - Cost Cost per attendee = 20 $$P(x) = R(x) - 20 \times A(x) = (20000 + 2000x)(100 - 5x) - 20(20000 + 2000x)$$ 11. **Expand profit:** $$P(x) = (20000 + 2000x)(100 - 5x) - 400,000 - 40,000x$$ From step 7, revenue part is: $$2,000,000 + 100,000x - 10,000x^2$$ So, $$P(x) = 2,000,000 + 100,000x - 10,000x^2 - 400,000 - 40,000x$$ $$= 1,600,000 + 60,000x - 10,000x^2$$ 12. **Maximize profit:** Take derivative and set to zero: $$\frac{dP}{dx} = 60,000 - 20,000x = 0$$ $$20,000x = 60,000$$ $$x = 3$$ 13. **Optimal ticket price for max profit:** $$p = 100 - 5 \times 3 = 85$$ **Final answers:** - Optimal ticket price for maximum revenue: $75$ - Optimal ticket price for maximum profit: $85$