1. **State the problem:** We want to find the price $P$ that maximizes the revenue function $$R = 2000 - 1000 (P - 2)^2.$$ 2. **Recall the rule for maximization:** A function is maximized where its derivative equals zero. The derivative given is $$R' = 2000 (P - 2).$$ 3. **Set the derivative equal to zero:** $$2000 (P - 2) = 0$$ 4. **Solve for $P$:** $$\cancel{2000} (P - 2) = \cancel{0}$$ $$P - 2 = 0$$ $$P = 2$$ 5. **Interpretation:** The price that maximizes revenue is $P = 2.00$. 6. **Check the options:** The correct choice is c) 2.00. **Final answer:** The price that maximizes revenue is **2.00**.