1. **Problem Statement:** We have a total cost function $$TC = 500 + 40Q - 0.5Q^2$$ and a total revenue function $$TR = 60Q$$. We need to find: a) The profit function $$\pi$$. b) The output level $$Q$$ that maximizes profit. c) The maximum profit or loss. 2. **Formulas and Rules:** - Profit function is $$\pi = TR - TC$$. - To maximize profit, find $$Q$$ where the derivative $$\frac{d\pi}{dQ} = 0$$. - For quadratic equations $$ax^2 + bx + c = 0$$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Step a) Find the profit function:** $$\pi = TR - TC = 60Q - (500 + 40Q - 0.5Q^2)$$ Simplify: $$\pi = 60Q - 500 - 40Q + 0.5Q^2 = 0.5Q^2 + 20Q - 500$$ 4. **Step b) Find profit-maximizing output:** Take derivative: $$\frac{d\pi}{dQ} = 2 \times 0.5 Q + 20 = Q + 20$$ Set derivative to zero for maximum: $$Q + 20 = 0 \implies Q = -20$$ Since negative output is not feasible, check if this is a maximum or minimum by second derivative: $$\frac{d^2\pi}{dQ^2} = 1 > 0$$, so this is a minimum. Because the profit function opens upward (positive coefficient on $$Q^2$$), the maximum profit occurs at the boundary or vertex. 5. **Find vertex of profit function:** Vertex $$Q = -\frac{b}{2a} = -\frac{20}{2 \times 0.5} = -\frac{20}{1} = -20$$ again negative, so no positive vertex. 6. **Check profit at $$Q=0$$:** $$\pi(0) = 0.5(0)^2 + 20(0) - 500 = -500$$ (loss) 7. **Check profit at some positive values:** Try $$Q=40$$: $$\pi(40) = 0.5(1600) + 20(40) - 500 = 800 + 800 - 500 = 1100$$ profit Try $$Q=60$$: $$\pi(60) = 0.5(3600) + 20(60) - 500 = 1800 + 1200 - 500 = 2500$$ profit Try $$Q=80$$: $$\pi(80) = 0.5(6400) + 20(80) - 500 = 3200 + 1600 - 500 = 4300$$ profit Since profit increases with $$Q$$, but the cost function has a negative quadratic term, re-examine the original cost function and profit function. **Note:** The cost function is $$TC = 500 + 40Q - 0.5Q^2$$, which decreases marginal cost initially but the negative quadratic term means cost decreases as $$Q$$ increases, which is unusual. **Recalculate derivative of profit:** $$\pi = TR - TC = 60Q - (500 + 40Q - 0.5Q^2) = 0.5Q^2 + 20Q - 500$$ Derivative: $$\frac{d\pi}{dQ} = Q + 20$$ Set to zero: $$Q = -20$$ (not feasible) Since $$\frac{d\pi}{dQ} = Q + 20$$ is positive for all $$Q > -20$$, profit increases as $$Q$$ increases. Therefore, profit increases without bound as $$Q$$ increases, which is unrealistic. **Conclusion:** The problem likely expects us to find maximum profit within realistic production limits or re-check the cost function. **Assuming the cost function is $$TC = 500 + 40Q + 0.5Q^2$$ (positive quadratic term), then:** Profit function: $$\pi = 60Q - (500 + 40Q + 0.5Q^2) = -0.5Q^2 + 20Q - 500$$ Derivative: $$\frac{d\pi}{dQ} = -Q + 20$$ Set to zero: $$-Q + 20 = 0 \implies Q = 20$$ Second derivative: $$\frac{d^2\pi}{dQ^2} = -1 < 0$$ (maximum) 8. **Step c) Calculate maximum profit:** $$\pi(20) = -0.5(400) + 20(20) - 500 = -200 + 400 - 500 = -300$$ This is a loss, so maximum profit is negative, meaning the firm incurs a loss at all production levels. **Final answers:** - Profit function: $$\pi = -0.5Q^2 + 20Q - 500$$ - Profit-maximizing output: $$Q = 20$$ - Maximum profit (loss): $$-300$$ --- **Note:** If the original cost function is correct as given, the profit function increases without bound, which is unrealistic. The corrected cost function with positive quadratic term is used here for meaningful results.