Question: 1. A manufacturer can produce at most 60 units of a certain product each year. The demand equation for the product is $p = q + 50q - 800$ and his average function is given as: $\bar{c} = \frac{2}{3} q - 20 + \frac{5000}{q}$. Determine the profit maximizing output approximated to nearest absolute WHOLE number

1. **State the problem:** We want to find the output $q$ that maximizes the profit for a manufacturer who can produce at most 60 units per year. 2. **Given:** - Demand equation: $$p = q + 50q - 800$$ - Average cost function: $$\bar{c} = \frac{2}{3} q - 20 + \frac{5000}{q}$$ - Maximum production: 60 units 3. **Simplify the demand equation:** $$p = q + 50q - 800 = 51q - 800$$ 4. **Find total revenue (TR):** $$TR = p \times q = (51q - 800)q = 51q^2 - 800q$$ 5. **Find total cost (TC):** Average cost $\bar{c} = \frac{TC}{q}$, so $$TC = \bar{c} \times q = \left(\frac{2}{3} q - 20 + \frac{5000}{q}\right) q = \frac{2}{3} q^2 - 20q + 5000$$ 6. **Profit function ($\pi$):** $$\pi = TR - TC = (51q^2 - 800q) - \left(\frac{2}{3} q^2 - 20q + 5000\right)$$ $$= 51q^2 - 800q - \frac{2}{3} q^2 + 20q - 5000$$ 7. **Simplify profit function:** $$\pi = \left(51 - \frac{2}{3}\right) q^2 + (-800 + 20) q - 5000$$ $$= \frac{153}{3} q^2 - \frac{2}{3} q^2 - 780 q - 5000 = \frac{151}{3} q^2 - 780 q - 5000$$ 8. **Find critical points by differentiating profit and setting to zero:** $$\frac{d\pi}{dq} = 2 \times \frac{151}{3} q - 780 = \frac{302}{3} q - 780$$ Set derivative to zero: $$\frac{302}{3} q - 780 = 0$$ 9. **Solve for $q$:** $$\frac{302}{3} q = 780$$ $$q = \frac{780 \times 3}{302} = \frac{2340}{302} \approx 7.75$$ 10. **Check if $q=7.75$ is within production limit:** Yes, $7.75 \leq 60$. 11. **Check second derivative to confirm maximum:** $$\frac{d^2 \pi}{dq^2} = 2 \times \frac{151}{3} = \frac{302}{3} > 0$$ Since second derivative is positive, this is a minimum, not maximum. 12. **Check profit at boundaries $q=0$ and $q=60$:** - At $q=0$, profit is undefined or zero. - At $q=60$: $$\pi(60) = \frac{151}{3} \times 60^2 - 780 \times 60 - 5000$$ Calculate: $$60^2 = 3600$$ $$\frac{151}{3} \times 3600 = 151 \times 1200 = 181200$$ $$-780 \times 60 = -46800$$ $$\pi(60) = 181200 - 46800 - 5000 = 129400$$ 13. **Since profit increases with $q$ and critical point is minimum, maximum profit is at $q=60$.** 14. **Final answer:** The profit maximizing output is approximately $$\boxed{60}$$ units.