Let's solve the problem step by step: --- ## **Step 1: Write Down Given Information** - **Cost function:** \[ C(q, K) = \frac{25q^2}{K} + 15K \] - **Marginal Cost (MC):** \[ MC(q, K) = \frac{50q}{K} \] - **K (robot hours):** 10 - **Price per sprocket (P):** $250 --- ## **Step 2: Plug in K = 10** ### Cost function: \[ C(q, 10) = \frac{25q^2}{10} + 15 \times 10 = 2.5q^2 + 150 \] ### Marginal cost: \[ MC(q, 10) = \frac{50q}{10} = 5q \] --- ## **Step 3: Find Profit-Maximizing Output** Profit is maximized where **Marginal Cost = Price**: \[ MC(q, 10) = P \] \[ 5q = 250 \] \[ q^* = \frac{250}{5} = 50 \] --- ## **Step 4: Calculate Profits** Profit (\(\pi\)) = Total Revenue - Total Cost - **Total Revenue:** \(TR = P \times q = 250 \times 50 = 12{,}500\) - **Total Cost:** \[ C(50, 10) = 2.5(50^2) + 150 = 2.5(250) + 150 = 625 + 150 = 640 \] So, \[ \pi = TR - TC = 12{,}500 - 6{,}400 = 6{,}100 \] --- ## **Step 5: Final Answers** ### **Profit-maximizing output level:** \[ \boxed{50 \text{ sprockets}} \] ### **Spacely's profits:** \[ \boxed{\$6{,}100} \] --- ### **Summary Table** | Output (\(q^*\)) | Profit (\(\pi\)) | |------------------|-----------------| | 50 sprockets | $6,100 |