1. **State the problem:** We are given a price-demand function $P(x) = 100 - x$ and a cost function $C(x) = 30000 + 50x$. We need to: a) Write the revenue function $R(x) = x \cdot P(x)$. b) Find the marginal cost function $C'(x)$. c) Find the marginal revenue function $R'(x)$. d) Calculate $R'(1000)$ and interpret its meaning. 2. **Write the revenue function:** Revenue is price times quantity sold, so $$R(x) = x \cdot P(x) = x(100 - x)$$ Expanding, $$R(x) = 100x - x^2$$ 3. **Find the marginal cost function $C'(x)$:** The cost function is $$C(x) = 30000 + 50x$$ The marginal cost is the derivative of cost with respect to $x$: $$C'(x) = \frac{d}{dx}(30000 + 50x) = 0 + 50 = 50$$ This means the cost increases by 50 for each additional item produced. 4. **Find the marginal revenue function $R'(x)$:** Given $$R(x) = 100x - x^2$$ Taking derivative, $$R'(x) = \frac{d}{dx}(100x - x^2) = 100 - 2x$$ 5. **Calculate $R'(1000)$:** $$R'(1000) = 100 - 2(1000) = 100 - 2000 = -1900$$ 6. **Interpretation:** $R'(1000) = -1900$ means that producing the 1001st item will decrease revenue by 1900 units of currency. In practical terms, producing one more item beyond 1000 reduces revenue, so it is not profitable to produce more than 1000 items.