1. **Problem Statement:** We have the price-demand equation $x = -30p + 9000$ and the cost function $C(x) = 30x + 150000$, where $x$ is the number of tables sold and $p$ is the price per table. 2. **Part a: Express $p$ as a function of $x$** From $x = -30p + 9000$, solve for $p$: $$x = -30p + 9000 \implies -30p = x - 9000 \implies p = \frac{9000 - x}{30}$$ 3. **Part b: Find the revenue function $R(x)$ and marginal revenue $R'(x)$** Revenue is price times quantity: $$R(x) = p \times x = \frac{9000 - x}{30} \times x = \frac{9000x - x^2}{30} = 300x - \frac{x^2}{30}$$ Marginal revenue is the derivative: $$R'(x) = \frac{d}{dx} \left(300x - \frac{x^2}{30}\right) = 300 - \frac{2x}{30} = 300 - \frac{x}{15}$$ 4. **Part c: Find $R'(3000)$ and $R'(6000)$ and interpret** Calculate: $$R'(3000) = 300 - \frac{3000}{15} = 300 - 200 = 100$$ $$R'(6000) = 300 - \frac{6000}{15} = 300 - 400 = -100$$ Interpretation: At $x=3000$, revenue is increasing at a rate of 100 per additional table sold. At $x=6000$, revenue is decreasing at a rate of 100 per additional table sold. 5. **Part d: Is $R(x)$ concave upward?** Second derivative: $$R''(x) = \frac{d}{dx} R'(x) = \frac{d}{dx} \left(300 - \frac{x}{15}\right) = -\frac{1}{15} < 0$$ Since $R''(x) < 0$, $R(x)$ is concave downward, not upward. 6. **Part e: Find profit function $P(x)$ and marginal profit $P'(x)$** Profit is revenue minus cost: $$P(x) = R(x) - C(x) = \left(300x - \frac{x^2}{30}\right) - (30x + 150000) = 300x - \frac{x^2}{30} - 30x - 150000 = 270x - \frac{x^2}{30} - 150000$$ Marginal profit: $$P'(x) = \frac{d}{dx} P(x) = 270 - \frac{2x}{30} = 270 - \frac{x}{15}$$ 7. **Part f: Approximate profit from selling the 55th unit** Use marginal profit at $x=54$ (approximate profit increase from 54 to 55 units): $$P'(54) = 270 - \frac{54}{15} = 270 - 3.6 = 266.4$$ Approximate profit from 55th unit is about 266.4. 8. **Part g: Exact profit from selling the 55th unit** Exact profit from 55th unit is: $$P(55) - P(54)$$ Calculate: $$P(55) = 270(55) - \frac{55^2}{30} - 150000 = 14850 - \frac{3025}{30} - 150000 = 14850 - 100.8333 - 150000 = -135250.8333$$ $$P(54) = 270(54) - \frac{54^2}{30} - 150000 = 14580 - \frac{2916}{30} - 150000 = 14580 - 97.2 - 150000 = -135517.2$$ Difference: $$P(55) - P(54) = -135250.8333 - (-135517.2) = 266.3667$$ 9. **Part h: Find number of tables sold that maximizes profit** Set marginal profit to zero: $$P'(x) = 270 - \frac{x}{15} = 0 \implies \frac{x}{15} = 270 \implies x = 4050$$ Second derivative of profit is same as revenue's second derivative: $$P''(x) = -\frac{1}{15} < 0$$ So $x=4050$ maximizes profit. **Final answers:** - $p(x) = \frac{9000 - x}{30}$ - $R(x) = 300x - \frac{x^2}{30}$, $R'(x) = 300 - \frac{x}{15}$ - $R'(3000) = 100$, $R'(6000) = -100$ - $R(x)$ is concave downward - $P(x) = 270x - \frac{x^2}{30} - 150000$, $P'(x) = 270 - \frac{x}{15}$ - Approximate profit from 55th unit: 266.4 - Exact profit from 55th unit: 266.37 - Profit maximized at $x=4050$ tables sold