1. **State the problem:** We are given the demand equation $q + 70p = 1400$ and the total cost function $TC(q) = 120 + 9q + q^2$. We want to find the quantity $q$ that maximizes the firm's profit. 2. **Recall the profit formula:** Profit $\pi(q)$ is revenue minus cost. 3. **Find the price function $p(q)$:** From the demand equation, $$q + 70p = 1400 \implies 70p = 1400 - q \implies p = \frac{1400 - q}{70} = 20 - \frac{q}{70}$$ 4. **Write the revenue function $R(q)$:** $$R(q) = p \times q = \left(20 - \frac{q}{70}\right) q = 20q - \frac{q^2}{70}$$ 5. **Write the profit function $\pi(q)$:** $$\pi(q) = R(q) - TC(q) = \left(20q - \frac{q^2}{70}\right) - \left(120 + 9q + q^2\right) = 20q - \frac{q^2}{70} - 120 - 9q - q^2$$ Simplify: $$\pi(q) = (20q - 9q) - \left(\frac{q^2}{70} + q^2\right) - 120 = 11q - \left(\frac{q^2}{70} + \frac{70q^2}{70}\right) - 120 = 11q - \frac{71q^2}{70} - 120$$ 6. **Maximize profit by finding critical points:** Take derivative: $$\pi'(q) = 11 - \frac{142q}{70} = 11 - \frac{142q}{70}$$ Set derivative to zero: $$11 - \frac{142q}{70} = 0 \implies \frac{142q}{70} = 11 \implies q = \frac{11 \times 70}{142} = \frac{770}{142} = \frac{385}{71}$$ 7. **Simplify fraction:** $$\frac{385}{71} = \frac{55 \times 7}{71}$$ Since 71 is prime and does not divide 385, fraction stays as is. 8. **Approximate value:** $$\frac{385}{71} \approx 5.42$$ 9. **Compare with options:** Options given are $\frac{51}{2} = 25.5$, $\frac{53}{2} = 26.5$, 27, and $\frac{55}{2} = 27.5$. Our calculated $q \approx 5.42$ does not match any option, so re-check step 6 derivative calculation. 10. **Recalculate derivative carefully:** $$\pi(q) = 11q - \frac{71q^2}{70} - 120$$ Derivative: $$\pi'(q) = 11 - 2 \times \frac{71}{70} q = 11 - \frac{142}{70} q = 11 - \frac{71}{35} q$$ Set to zero: $$11 = \frac{71}{35} q \implies q = \frac{11 \times 35}{71} = \frac{385}{71} \approx 5.42$$ Still 5.42, no match. 11. **Check if profit maximization is at boundary or if options are incorrect:** Since options are much larger, check if price function or cost function was misread. 12. **Re-express price function:** From $q + 70p = 1400$, $p = \frac{1400 - q}{70} = 20 - \frac{q}{70}$ correct. 13. **Revenue:** $$R = p q = q \left(20 - \frac{q}{70}\right) = 20q - \frac{q^2}{70}$$ 14. **Profit:** $$\pi = R - TC = 20q - \frac{q^2}{70} - (120 + 9q + q^2) = 20q - \frac{q^2}{70} - 120 - 9q - q^2 = 11q - 120 - \left(q^2 + \frac{q^2}{70}\right)$$ Combine quadratic terms: $$q^2 + \frac{q^2}{70} = q^2 \left(1 + \frac{1}{70}\right) = q^2 \frac{71}{70}$$ So, $$\pi = 11q - 120 - \frac{71}{70} q^2$$ 15. **Derivative:** $$\pi' = 11 - 2 \times \frac{71}{70} q = 11 - \frac{142}{70} q = 11 - \frac{71}{35} q$$ Set to zero: $$11 = \frac{71}{35} q \implies q = \frac{11 \times 35}{71} = \frac{385}{71} \approx 5.42$$ 16. **Conclusion:** The profit is maximized at $q = \frac{385}{71}$, approximately 5.42, which does not match any given options. Since the problem asks to select from options, and none matches, the closest is none. **Final answer:** None of the given options is correct based on the calculations.