1. **Problem statement:** We are given the cost function $C(x) = 0.015x^2 + 10x + 300$ and the marginal revenue function $R'(x) = 60 - 0.02x$. We need to: a) Find the revenue function $R(x)$ and the change in total revenue when production increases from 60 to 70 units. b) Find the number of units that maximize profit and calculate the maximum profit. 2. **Find the revenue function $R(x)$:** The marginal revenue $R'(x)$ is the derivative of the revenue function $R(x)$. To find $R(x)$, integrate $R'(x)$: $$R(x) = \int (60 - 0.02x) \, dx = 60x - 0.01x^2 + C$$ where $C$ is the constant of integration. 3. **Determine the constant $C$:** Assuming no revenue when no units are sold, $R(0) = 0$: $$R(0) = 60 \times 0 - 0.01 \times 0^2 + C = C = 0$$ So, $$R(x) = 60x - 0.01x^2$$ 4. **Calculate the change in total revenue from 60 to 70 units:** $$\Delta R = R(70) - R(60)$$ Calculate each: $$R(70) = 60 \times 70 - 0.01 \times 70^2 = 4200 - 0.01 \times 4900 = 4200 - 49 = 4151$$ $$R(60) = 60 \times 60 - 0.01 \times 60^2 = 3600 - 0.01 \times 3600 = 3600 - 36 = 3564$$ So, $$\Delta R = 4151 - 3564 = 587$$ 5. **Find the profit function $P(x)$:** Profit is revenue minus cost: $$P(x) = R(x) - C(x) = (60x - 0.01x^2) - (0.015x^2 + 10x + 300)$$ Simplify: $$P(x) = 60x - 0.01x^2 - 0.015x^2 - 10x - 300 = (60x - 10x) - (0.01x^2 + 0.015x^2) - 300 = 50x - 0.025x^2 - 300$$ 6. **Find the number of units that maximize profit:** To maximize profit, find $x$ where the derivative $P'(x) = 0$: $$P'(x) = \frac{d}{dx}(50x - 0.025x^2 - 300) = 50 - 0.05x$$ Set derivative to zero: $$50 - 0.05x = 0$$ $$0.05x = 50$$ $$x = \frac{50}{0.05} = 1000$$ 7. **Calculate the maximum profit:** Substitute $x=1000$ into $P(x)$: $$P(1000) = 50 \times 1000 - 0.025 \times 1000^2 - 300 = 50000 - 0.025 \times 1000000 - 300 = 50000 - 25000 - 300 = 24700$$ **Final answers:** - Revenue function: $R(x) = 60x - 0.01x^2$ - Change in revenue from 60 to 70 units: $587$ - Units for maximum profit: $1000$ - Maximum profit: $24700$