1. **Problem Statement:** We want to find the number of air conditioning units $q$ that ZamFlow should produce each month to maximize profit. Given: - Cost function: $$C(q) = 7000q + 20q^2$$ - Revenue function: $$R(q) = -\frac{297}{4}q^2 + 29700q + 30000$$ 2. **Profit Function:** Profit $P(q)$ is revenue minus cost: $$P(q) = R(q) - C(q) = \left(-\frac{297}{4}q^2 + 29700q + 30000\right) - \left(7000q + 20q^2\right)$$ Simplify: $$P(q) = -\frac{297}{4}q^2 - 20q^2 + 29700q - 7000q + 30000$$ $$P(q) = -\left(\frac{297}{4} + 20\right)q^2 + (29700 - 7000)q + 30000$$ Calculate coefficients: $$\frac{297}{4} = 74.25$$ So, $$P(q) = -(74.25 + 20)q^2 + 22700q + 30000 = -94.25q^2 + 22700q + 30000$$ 3. **Maximizing Profit:** To find the maximum profit, take the derivative of $P(q)$ and set it to zero: $$P'(q) = -2 \times 94.25 q + 22700 = -188.5q + 22700$$ Set derivative to zero: $$-188.5q + 22700 = 0$$ $$188.5q = 22700$$ $$q = \frac{22700}{188.5} \approx 120.42$$ 4. **Second Derivative Test:** $$P''(q) = -188.5 < 0$$ Since $P''(q) < 0$, the critical point at $q \approx 120.42$ is a maximum. 5. **Calculate Maximum Profit:** Substitute $q = 120.42$ into $P(q)$: $$P(120.42) = -94.25(120.42)^2 + 22700(120.42) + 30000$$ Calculate: $$120.42^2 \approx 14500.3$$ $$-94.25 \times 14500.3 \approx -1366267.3$$ $$22700 \times 120.42 \approx 2737434$$ So, $$P(120.42) \approx -1366267.3 + 2737434 + 30000 = 1399166.7$$ 6. **Compare with Break-even Strategy:** At $q=242$ units: $$P(242) = -94.25(242)^2 + 22700(242) + 30000$$ Calculate: $$242^2 = 58564$$ $$-94.25 \times 58564 = -5519987$$ $$22700 \times 242 = 5493400$$ So, $$P(242) = -5519987 + 5493400 + 30000 = 80413$$ 7. **Summary:** - Profit-maximizing output: approximately 120 units per month. - Maximum profit: approximately 1,399,167 kwacha. - Current break-even output: 242 units with profit approximately 80,413 kwacha. - Changing to profit-maximizing output increases profit significantly but reduces output. **Slug:** profit maximization **Subject:** economics