Question: Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is $p$ dollars, the revenue $R$ (in dollars) is $$R(p) = -9p^2 + 18000p$$. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?

1. **State the problem:** We want to find the unit price $p$ that maximizes the revenue function $$R(p) = -9p^2 + 18000p$$ and then find the maximum revenue. 2. **Formula and rules:** The revenue function is a quadratic function of the form $$R(p) = ap^2 + bp + c$$ where $a = -9$, $b = 18000$, and $c = 0$. Since $a < 0$, the parabola opens downward, so the vertex represents the maximum point. 3. **Find the vertex:** The $p$-coordinate of the vertex is given by $$p = -\frac{b}{2a} = -\frac{18000}{2 \times (-9)} = -\frac{18000}{-18} = 1000$$. 4. **Calculate the maximum revenue:** Substitute $p=1000$ into the revenue function: $$R(1000) = -9(1000)^2 + 18000(1000) = -9 \times 1,000,000 + 18,000,000 = -9,000,000 + 18,000,000 = 9,000,000$$. 5. **Interpretation:** The unit price that maximizes revenue is $1000$, and the maximum revenue is $9,000,000$. **Final answers:** - Unit price to maximize revenue: $1000$ - Maximum revenue: $9,000,000$