1. **State the problem:** We have the average cost function $AC = \frac{80}{Q} + 2 + Q$ and a linear demand function with points $(Q=1, P=29)$ and $(Q=10, P=20)$. We need to find the total cost, total revenue, and profit functions. 2. **Find the demand (price) function:** The demand function is linear: $P = mQ + b$. Calculate slope $m$: $$m = \frac{20 - 29}{10 - 1} = \frac{-9}{9} = -1$$ Use point-slope form with point $(1,29)$: $$29 = -1 \times 1 + b \Rightarrow b = 30$$ So, $$P = -Q + 30$$ 3. **Find total cost (TC):** Average cost $AC = \frac{TC}{Q}$, so $$TC = AC \times Q = \left(\frac{80}{Q} + 2 + Q\right) Q = 80 + 2Q + Q^2$$ 4. **Find total revenue (TR):** $$TR = P \times Q = (-Q + 30) Q = -Q^2 + 30Q$$ 5. **Find profit function ($\pi$):** Profit is revenue minus cost: $$\pi = TR - TC = (-Q^2 + 30Q) - (80 + 2Q + Q^2)$$ Simplify: $$\pi = -Q^2 + 30Q - 80 - 2Q - Q^2 = -2Q^2 + 28Q - 80$$ 6. **Find values of $Q$ for profit ($\pi > 0$):** Solve $$-2Q^2 + 28Q - 80 > 0$$ Divide by -2 (reverse inequality): $$Q^2 - 14Q + 40 < 0$$ Find roots: $$Q = \frac{14 \pm \sqrt{14^2 - 4 \times 40}}{2} = \frac{14 \pm \sqrt{196 - 160}}{2} = \frac{14 \pm 6}{2}$$ Roots are $Q=4$ and $Q=10$. Since parabola opens upward, inequality $<0$ holds between roots: $$4 < Q < 10$$ 7. **Find values of $Q$ for loss ($\pi < 0$):** This is outside the interval where profit is positive: $$Q < 4 \quad \text{or} \quad Q > 10$$ **Final answers:** - Total cost: $$TC = 80 + 2Q + Q^2$$ - Total revenue: $$TR = -Q^2 + 30Q$$ - Profit function: $$\pi = -2Q^2 + 28Q - 80$$ - Profit when $$4 < Q < 10$$ - Loss when $$Q < 4$$ or $$Q > 10$$