1. **State the problem:** We have a profit function $\pi = -2Q^2 + 9Q - 4$. (a) Find the break-even values of $Q$ and the maximum profit. (b) Find the demand function given fixed costs of 4 and variable costs of 1 per unit. 2. **Break-even values:** Break-even occurs when profit $\pi = 0$. Set the profit function equal to zero: $$-2Q^2 + 9Q - 4 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula: $$Q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = -2$, $b = 9$, and $c = -4$. Calculate the discriminant: $$\Delta = 9^2 - 4(-2)(-4) = 81 - 32 = 49$$ Calculate the roots: $$Q = \frac{-9 \pm \sqrt{49}}{2(-2)} = \frac{-9 \pm 7}{-4}$$ Two solutions: - $$Q_1 = \frac{-9 + 7}{-4} = \frac{-2}{-4} = 0.5$$ - $$Q_2 = \frac{-9 - 7}{-4} = \frac{-16}{-4} = 4$$ So, break-even values are $Q = 0.5$ and $Q = 4$. 4. **Maximum profit:** Since the profit function is a downward-opening parabola ($a = -2 < 0$), the maximum profit occurs at the vertex. Vertex formula for $Q$: $$Q_{max} = -\frac{b}{2a} = -\frac{9}{2(-2)} = \frac{9}{4} = 2.25$$ Calculate maximum profit: $$\pi_{max} = -2(2.25)^2 + 9(2.25) - 4 = -2(5.0625) + 20.25 - 4 = -10.125 + 20.25 - 4 = 6.125$$ 5. **Find the demand function:** Profit is revenue minus cost: $$\pi = R - C$$ Revenue $R = P \times Q$, where $P$ is price. Cost $C = \text{fixed costs} + \text{variable costs} \times Q = 4 + 1 \times Q = 4 + Q$. Given profit function: $$\pi = -2Q^2 + 9Q - 4$$ Rewrite profit as: $$\pi = R - C = P Q - (4 + Q)$$ So, $$P Q - (4 + Q) = -2Q^2 + 9Q - 4$$ Simplify: $$P Q - 4 - Q = -2Q^2 + 9Q - 4$$ Add 4 to both sides: $$P Q - Q = -2Q^2 + 9Q$$ Factor left side: $$Q(P - 1) = -2Q^2 + 9Q$$ Divide both sides by $Q$ (assuming $Q \neq 0$): $$P - 1 = -2Q + 9$$ Solve for $P$: $$P = -2Q + 9 + 1 = -2Q + 10$$ **Demand function:** $$P = -2Q + 10$$ This means price decreases by 2 units for each additional unit sold. **Final answers:** - Break-even values: $Q = 0.5$ and $Q = 4$ - Maximum profit: $6.125$ at $Q = 2.25$ - Demand function: $P = -2Q + 10$