1. **Stating the problem:** ABCD Farms has a total cost function $$TC = 100 + 12Q + 0.06Q^2$$ where $Q$ represents 100 gallons of milk. We are to find cost functions, economies of scale, shutdown and break-even points, market supply, equilibrium, and profit. 2. **a. Cost functions:** i. Average Total Cost (ATC): $$ATC = \frac{TC}{Q} = \frac{100 + 12Q + 0.06Q^2}{Q} = \frac{100}{Q} + 12 + 0.06Q$$ ii. Average Variable Cost (AVC): Variable cost is $VC = 12Q + 0.06Q^2$, so $$AVC = \frac{VC}{Q} = 12 + 0.06Q$$ iii. Marginal Cost (MC): $$MC = \frac{dTC}{dQ} = 12 + 0.12Q$$ iv. Total Fixed Cost (TFC): The fixed cost is the constant term in $TC$, so $$TFC = 100$$ 3. **b. Economies or diseconomies of scale?** Economies of scale occur if ATC decreases as $Q$ increases. Since $$ATC = \frac{100}{Q} + 12 + 0.06Q,$$ the term $\frac{100}{Q}$ decreases but $0.06Q$ increases as $Q$ increases. To check where ATC is minimized, derive: $$\frac{d(ATC)}{dQ} = -\frac{100}{Q^2} + 0.06$$ Set to zero for minimum: $$-\frac{100}{Q^2} + 0.06 = 0 \implies \frac{100}{Q^2} = 0.06 \implies Q^2 = \frac{100}{0.06} = 1666.67 \implies Q \approx 40.82$$ For $Q < 40.82$, ATC decreases (economies of scale). For $Q > 40.82$, ATC increases (diseconomies of scale). 4. **c. Shutdown and break-even price and output levels:** Shutdown occurs where price equals minimum AVC. AVC is minimum where: $$\frac{d(AVC)}{dQ} = 0.06 = 0 > 0,$$ so AVC is increasing in $Q$; minimum AVC occurs at $Q=0$ with $$AVC(0) = 12.$$ So shutdown price $$P_{shutdown} = 12.$$ Output at shutdown is $0$. Break-even occurs where price equals minimum ATC. As calculated, minimum ATC at $Q \approx 40.82$. Find minimum ATC: $$ATC_{min} = \frac{100}{40.82} + 12 + 0.06 \times 40.82 \approx 2.45 + 12 + 2.45 = 16.9$$ So, $$P_{break-even} = 16.9, \quad Q_{break-even} = 40.82.$$ 5. **d. Market supply curve:** Each firm’s supply curve is the portion of MC above AVC (since price must cover AVC to supply). Since MC = 12 + 0.12Q and AVC = 12 + 0.06Q, MC > AVC for all $Q > 0$. Thus supply curve is $$P = 12 + 0.12Q \implies Q = \frac{P - 12}{0.12} = 8.33(P - 12)$$ For one firm. With 5,000 identical firms, $$Q_s = 5000 \times 8.33 (P - 12) = 41667 (P - 12)$$ 6. **e. Market equilibrium price and quantity:** Market demand: $$Q_D = 660,000 - 16,333.33 P$$ Market supply: $$Q_s = 41,667 (P - 12)$$ At equilibrium: $$Q_D = Q_s$$ $$660,000 - 16,333.33 P = 41,667 (P - 12)$$ Solve: $$660,000 - 16,333.33 P = 41,667 P - 500,004$$ $$660,000 + 500,004 = 41,667 P + 16,333.33 P$$ $$1,160,004 = 58,000.33 P$$ $$P = \frac{1,160,004}{58,000.33} \approx 20$$ Substitute $P=20$ into $Q_s$: $$Q = 41,667 (20 - 12) = 41,667 \times 8 = 333,336$$ 7. **f. ABCD farm’s profit:** Recall $Q$ is in units of 100 gallons. So farm’s output at $P=20$: $$Q = \frac{P - 12}{0.12} = \frac{20 - 12}{0.12} = 66.67$$ (units of 100 gallons) Revenue: $$TR = P \times Q = 20 \times 66.67 = 1,333.4$$ (in hundreds of gallons and dollars per hundred gallons) Total Cost: $$TC = 100 + 12 \times 66.67 + 0.06 \times (66.67)^2 = 100 + 800 + 0.06 \times 4,444.89 = 100 + 800 + 266.69 = 1,166.69$$ Profit: $$\pi = TR - TC = 1,333.4 - 1,166.69 = 166.71$$