1. **State the problem:** We want to find the production level $x$ that maximizes the profit, given the revenue function $R(x) = 6x$ and the cost function $C(x) = 0.05x^2 + 0.7x + 8$. 2. **Formula for profit:** Profit $P(x)$ is revenue minus cost: $$P(x) = R(x) - C(x) = 6x - (0.05x^2 + 0.7x + 8)$$ 3. **Simplify the profit function:** $$P(x) = 6x - 0.05x^2 - 0.7x - 8 = -0.05x^2 + 5.3x - 8$$ 4. **Find the maximum profit:** Since $P(x)$ is a quadratic function with a negative leading coefficient, it opens downward and has a maximum at its vertex. 5. **Vertex formula:** The $x$-value of the vertex is given by $$x = -\frac{b}{2a}$$ where $a = -0.05$ and $b = 5.3$. 6. **Calculate $x$ for maximum profit:** $$x = -\frac{5.3}{2 \times (-0.05)} = -\frac{5.3}{-0.1} = 53$$ 7. **Calculate the maximum profit:** Substitute $x=53$ into $P(x)$: $$P(53) = -0.05(53)^2 + 5.3(53) - 8$$ $$= -0.05 \times 2809 + 280.9 - 8$$ $$= -140.45 + 280.9 - 8 = 132.45$$ **Final answers:** - Maximum profit occurs when producing and selling **53 units**. - The maximum profit is **132.45 dollars**.