1. **State the problem:** We want to find the price per phone $x$ that maximizes the revenue function $$r(x) = x \cdot (1200 - 2x)$$ where $x$ is the price per phone in dollars. 2. **Write the revenue function in standard quadratic form:** $$r(x) = x(1200 - 2x) = 1200x - 2x^2$$ 3. **Identify coefficients:** The quadratic function is $$r(x) = -2x^2 + 1200x + 0$$ where $a = -2$, $b = 1200$, and $c = 0$. 4. **Find the vertex (maximum point) of the parabola:** Since $a < 0$, the parabola opens downward and the vertex gives the maximum revenue. The $x$-coordinate of the vertex is given by the formula: $$x = -\frac{b}{2a} = -\frac{1200}{2 \times (-2)} = -\frac{1200}{-4} = 300$$ 5. **Calculate the maximum revenue by substituting $x=300$ into $r(x)$:** $$r(300) = 300 \times (1200 - 2 \times 300) = 300 \times (1200 - 600) = 300 \times 600 = 180000$$ 6. **Interpretation:** The maximum revenue of 180,000 dollars occurs when the price per phone is 300 dollars. **Final answer:** The price per phone that maximizes revenue is $\boxed{300}$ dollars, and the maximum revenue is $\boxed{180000}$ dollars.