1. **State the problem:** We are given the cost function $C(x) = 0.5x^2 + 50x + 100$ and the revenue function $R(x) = 80x - 0.01x^2$. We need to find the profit function $P(x)$, then use calculus to find the production level $x$ that maximizes profit and the maximum profit value. 2. **Find the profit function:** Profit is revenue minus cost: $$P(x) = R(x) - C(x)$$ Substitute the given functions: $$P(x) = (80x - 0.01x^2) - (0.5x^2 + 50x + 100)$$ Simplify: $$P(x) = 80x - 0.01x^2 - 0.5x^2 - 50x - 100$$ Combine like terms: $$P(x) = (80x - 50x) - (0.01x^2 + 0.5x^2) - 100$$ $$P(x) = 30x - 0.51x^2 - 100$$ Rearranged: $$P(x) = -0.51x^2 + 30x - 100$$ This matches the required form. 3. **Maximize the profit using calculus:** To find the maximum profit, differentiate $P(x)$ with respect to $x$: $$P'(x) = \frac{d}{dx}(-0.51x^2 + 30x - 100) = -1.02x + 30$$ Set derivative equal to zero to find critical points: $$-1.02x + 30 = 0$$ $$-1.02x = -30$$ $$x = \frac{-30}{\cancel{-1.02}} \cancel{\times} \frac{1}{1.02} = \frac{30}{1.02}$$ Calculate: $$x \approx 29.41$$ Rounded to nearest whole number: $$x = 29$$ 4. **Find the maximum profit:** Substitute $x=29$ into $P(x)$: $$P(29) = -0.51(29)^2 + 30(29) - 100$$ Calculate $29^2$: $$29^2 = 841$$ So: $$P(29) = -0.51 \times 841 + 870 - 100$$ $$P(29) = -428.91 + 870 - 100$$ $$P(29) = 341.09$$ Rounded to nearest cent: $$P(29) = 341.09$$ **Final answers:** - Profit function: $P(x) = -0.51x^2 + 30x - 100$ - Production level that maximizes profit: $29$ units - Maximum profit: $341.09$