1. **State the problem:** We are given the price function $p = 80 - 0.01x$ and the cost function $C = 20x + 5000$. We need to find the production level $x$ that maximizes profit. 2. **Formula for profit:** Profit $P$ is revenue minus cost. Revenue $R$ is price times quantity, so: $$P(x) = R(x) - C(x) = p \times x - C(x) = (80 - 0.01x)x - (20x + 5000)$$ 3. **Express profit function:** $$P(x) = 80x - 0.01x^2 - 20x - 5000 = (80x - 20x) - 0.01x^2 - 5000 = 60x - 0.01x^2 - 5000$$ 4. **Find critical points:** To maximize profit, take derivative and set to zero: $$P'(x) = 60 - 0.02x$$ Set $P'(x) = 0$: $$60 - 0.02x = 0$$ $$0.02x = 60$$ $$x = \frac{60}{0.02}$$ $$x = 3000$$ 5. **Check second derivative to confirm maximum:** $$P''(x) = -0.02 < 0$$ Since $P''(x) < 0$, $x=3000$ is a maximum. 6. **Conclusion:** The production level for maximum profit is $\boxed{3000}$ units. Among the options, the correct answer is d) $x=3000$.