1. Let the Numbers Be \( x \) and \( y \): Given that \( x + y = 9 \), we can express \( y \) as: \[ y = 9 - x \] 2. Define the Function to Maximize: We need to maximize the product \( P = x y^2 \). Substitute \( y = 9 - x \): \[ P = x (9 - x)^2 \] 3. Expand the Function: \[ P = x (81 - 18x + x^2) = 81x - 18x^2 + x^3 \] 4. Find the Critical Points: To find the maximum, take the derivative of \( P \) with respect to \( x \) and set it to zero: \[ \frac{dP}{dx} = 81 - 36x + 3x^2 \] Set the derivative equal to zero: \[ 3x^2 - 36x + 81 = 0 \] Divide by 3: \[ x^2 - 12x + 27 = 0 \] Solve the quadratic equation: \[ x = \frac{12 \pm \sqrt{144 - 108}}{2} = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2} \] \[ x = 9 \quad \text{or} \quad x = 3 \] 5. Determine the Maximum: - If \( x = 9 \), then \( y = 0 \), and \( P = 9 \times 0^2 = 0 \). - If \( x = 3 \), then \( y = 6 \), and \( P = 3 \times 6^2 = 108 \). The maximum product occurs when \( x = 3 \) and \( y = 6 \). Therefore, the two numbers are: \[ \boxed{3 \text{ and } 6} \]