1. **State the problem:** We have the revenue function $$R(x) = -0.025x^2 + 900x$$ and the cost function $$C(x) = 200x + 280000$$. We need to: - Find the marginal revenue function $$R'(x)$$. - Find and interpret $$R'(5000)$$. - Find the profit function $$P(x)$$. - Find the marginal profit function $$P'(x)$$ and interpret $$P'(5000)$$. 2. **Find the marginal revenue function:** The marginal revenue function is the derivative of the revenue function with respect to $$x$$. Given $$R(x) = -0.025x^2 + 900x$$, $$R'(x) = \frac{d}{dx}(-0.025x^2 + 900x) = -0.05x + 900$$. 3. **Find and interpret $$R'(5000)$$:** Calculate: $$R'(5000) = -0.05(5000) + 900 = -250 + 900 = 650$$. Interpretation: At a production level of 5000 units, the marginal revenue is 650. This means producing one more unit around 5000 units will increase revenue by approximately 650. 4. **Find the profit function $$P(x)$$:** Profit is revenue minus cost: $$P(x) = R(x) - C(x) = (-0.025x^2 + 900x) - (200x + 280000)$$ Simplify: $$P(x) = -0.025x^2 + 900x - 200x - 280000 = -0.025x^2 + 700x - 280000$$. 5. **Find the marginal profit function $$P'(x)$$:** Take the derivative of $$P(x)$$: $$P'(x) = \frac{d}{dx}(-0.025x^2 + 700x - 280000) = -0.05x + 700$$. 6. **Find and interpret $$P'(5000)$$:** Calculate: $$P'(5000) = -0.05(5000) + 700 = -250 + 700 = 450$$. Interpretation: At 5000 units, the marginal profit is 450, meaning producing one more unit will increase profit by approximately 450.