1. **State the problem:** We are given the total revenue function $$R = 3x^2 + 200x$$ and the total cost function $$C = 2x^2 - 150x + 5000$$ where $x$ is the number of units sold in thousands. We need to find the number of units $x$ that maximizes the total profit. 2. **Define the profit function:** Profit $$P(x)$$ is revenue minus cost: $$P(x) = R(x) - C(x) = (3x^2 + 200x) - (2x^2 - 150x + 5000)$$ 3. **Simplify the profit function:** $$P(x) = 3x^2 + 200x - 2x^2 + 150x - 5000 = (3x^2 - 2x^2) + (200x + 150x) - 5000 = x^2 + 350x - 5000$$ 4. **Find the critical points:** To maximize profit, find where the derivative $$P'(x)$$ equals zero. $$P'(x) = \frac{d}{dx}(x^2 + 350x - 5000) = 2x + 350$$ Set $$P'(x) = 0$$: $$2x + 350 = 0$$ $$2x = -350$$ $$x = -175$$ 5. **Check the nature of critical point:** The second derivative is $$P''(x) = \frac{d}{dx}(2x + 350) = 2$$ Since $$P''(x) = 2 > 0$$, the function has a minimum at $$x = -175$$, not a maximum. 6. **Analyze the profit function:** Since the coefficient of $$x^2$$ in $$P(x)$$ is positive (1), the parabola opens upwards, so profit has no maximum but a minimum at $$x = -175$$. 7. **Interpretation:** Because $$x$$ represents thousands of units sold, negative units are not possible. The profit function increases without bound as $$x$$ increases. 8. **Conclusion:** The profit increases as $$x$$ increases, so to maximize profit, produce and sell as many units as possible within practical constraints. **Final answer:** There is no finite maximum profit for positive $$x$$; profit increases with the number of units sold. The minimum profit occurs at $$x = 0$$ or near zero units, and profit grows as $$x$$ increases.