1. **State the problem:** We have cost function $C(X) = 12X + 300$ and revenue function $R(X) = 0.4X^2 + 48X + 200$, where $X$ is in thousands of tickets. We need to: - Find the number of tickets $X$ that maximizes revenue. - Find the maximum revenue. - Obtain the profit function $P(X)$. - Find the breakeven point less than 20,000 units (i.e., $X < 20$). 2. **Maximizing revenue:** Revenue function is $R(X) = 0.4X^2 + 48X + 200$. Since the coefficient of $X^2$ is positive ($0.4 > 0$), this is a parabola opening upwards, so it has a minimum, not a maximum. Check if the problem meant to maximize revenue or profit. Usually, revenue is maximized if parabola opens downward (negative $X^2$ coefficient). Here, $0.4X^2$ is positive, so revenue increases without bound as $X$ increases. Assuming the problem wants to maximize profit instead, let's proceed to find profit function and maximize it. 3. **Profit function:** Profit $P(X) = R(X) - C(X)$ $$P(X) = (0.4X^2 + 48X + 200) - (12X + 300)$$ $$P(X) = 0.4X^2 + 48X + 200 - 12X - 300$$ $$P(X) = 0.4X^2 + (48 - 12)X + (200 - 300)$$ $$P(X) = 0.4X^2 + 36X - 100$$ 4. **Maximize profit:** Since $0.4 > 0$, $P(X)$ is a parabola opening upwards, so it has a minimum, not a maximum. This suggests profit increases without bound as $X$ increases, which is unusual. 5. **Check revenue function again:** If the revenue function is $R(X) = -0.4X^2 + 48X + 200$ (negative coefficient), then it opens downward and has a maximum. Assuming a typo and revenue is $R(X) = -0.4X^2 + 48X + 200$, let's proceed. 6. **Maximize revenue with corrected $R(X)$:** $$R(X) = -0.4X^2 + 48X + 200$$ Take derivative: $$R'(X) = -0.8X + 48$$ Set derivative to zero for max: $$-0.8X + 48 = 0$$ $$0.8X = 48$$ $$X = \frac{48}{0.8} = 60$$ 7. **Maximum revenue:** $$R(60) = -0.4(60)^2 + 48(60) + 200$$ $$= -0.4(3600) + 2880 + 200$$ $$= -1440 + 2880 + 200 = 1640$$ Maximum revenue is 1640 thousand euros at 60 thousand tickets. 8. **Profit function with corrected $R(X)$:** $$P(X) = R(X) - C(X) = (-0.4X^2 + 48X + 200) - (12X + 300)$$ $$P(X) = -0.4X^2 + 48X + 200 - 12X - 300$$ $$P(X) = -0.4X^2 + 36X - 100$$ 9. **Breakeven point:** Breakeven means $P(X) = 0$ $$-0.4X^2 + 36X - 100 = 0$$ Multiply both sides by $-1$: $$0.4X^2 - 36X + 100 = 0$$ Divide by 0.4: $$\cancel{0.4}X^2 - \cancel{36}X + \cancel{100} = 0 \Rightarrow X^2 - 90X + 250 = 0$$ Use quadratic formula: $$X = \frac{90 \pm \sqrt{90^2 - 4 \times 1 \times 250}}{2}$$ $$= \frac{90 \pm \sqrt{8100 - 1000}}{2} = \frac{90 \pm \sqrt{7100}}{2}$$ $$\sqrt{7100} \approx 84.26$$ So, $$X_1 = \frac{90 - 84.26}{2} = \frac{5.74}{2} = 2.87$$ $$X_2 = \frac{90 + 84.26}{2} = \frac{174.26}{2} = 87.13$$ Since breakeven point is less than 20,000 units, the breakeven point is at $X = 2.87$ thousand tickets. **Final answers:** - Number of tickets maximizing revenue: $60$ thousand - Maximum revenue: $1640$ thousand euros - Profit function: $$P(X) = -0.4X^2 + 36X - 100$$ - Breakeven point less than 20,000 units: $2.87$ thousand tickets