1. **State the problem:** We are given the marginal cost function $MC = 2000 - 320x + 3x^2$, fixed cost $= 18000$, and price per unit $= 6800$. We need to find the cost function, revenue function, profit function, sales volume for maximum profit, and the profit at that sales volume. 2. **Find the cost function $C(x)$:** Marginal cost is the derivative of the cost function: $MC = C'(x) = 2000 - 320x + 3x^2$. To find $C(x)$, integrate $MC$ with respect to $x$: $$C(x) = \int (2000 - 320x + 3x^2) dx = 2000x - 160x^2 + x^3 + K$$ Given fixed cost $C(0) = 18000$, so: $$C(0) = K = 18000$$ Thus, $$C(x) = 2000x - 160x^2 + x^3 + 18000$$ 3. **Find the revenue function $R(x)$:** Revenue is price per unit times units sold: $$R(x) = 6800x$$ 4. **Find the profit function $P(x)$:** Profit is revenue minus cost: $$P(x) = R(x) - C(x) = 6800x - (2000x - 160x^2 + x^3 + 18000)$$ Simplify: $$P(x) = 6800x - 2000x + 160x^2 - x^3 - 18000 = 4800x + 160x^2 - x^3 - 18000$$ 5. **Find sales volume $x$ that yields maximum profit:** To maximize profit, set derivative $P'(x)$ to zero: $$P'(x) = 4800 + 320x - 3x^2 = 0$$ Rewrite: $$-3x^2 + 320x + 4800 = 0$$ Multiply both sides by $-1$: $$3x^2 - 320x - 4800 = 0$$ Use quadratic formula: $$x = \frac{320 \pm \sqrt{(-320)^2 - 4 \times 3 \times (-4800)}}{2 \times 3} = \frac{320 \pm \sqrt{102400 + 57600}}{6} = \frac{320 \pm \sqrt{160000}}{6} = \frac{320 \pm 400}{6}$$ Two solutions: $$x_1 = \frac{320 + 400}{6} = \frac{720}{6} = 120$$ $$x_2 = \frac{320 - 400}{6} = \frac{-80}{6} = -13.33$$ Since sales volume cannot be negative, take $x = 120$ units. 6. **Find profit at $x=120$:** $$P(120) = 4800(120) + 160(120)^2 - (120)^3 - 18000$$ Calculate stepwise: $$4800 \times 120 = 576000$$ $$160 \times 14400 = 2304000$$ $$120^3 = 1728000$$ So, $$P(120) = 576000 + 2304000 - 1728000 - 18000 = 576000 + 2304000 - 1746000 = 1098000$$ **Final answers:** - Cost function: $C(x) = 2000x - 160x^2 + x^3 + 18000$ - Revenue function: $R(x) = 6800x$ - Profit function: $P(x) = 4800x + 160x^2 - x^3 - 18000$ - Sales volume for max profit: $x = 120$ - Maximum profit: $P(120) = 1098000$