Suppose the profit from the sale of \( x \) units of a product is \( P=6400 x_{2}-18 x^{2}-400 \). (a) What level(s) of production will yield a profit of \( \$ 318,800 \) ? (Enter your answers as a comma-separated list. Round your answers to two decimal places.) (b) Can a profit of more than \( \$ 318,800 \) be made? (Yes No

Let's analyze the given profit function: \[ P = 6400x - 18x^2 - 400 \] ### Part (a): Determining Production Levels for a Profit of \$318,800 We need to find the values of \( x \) such that \( P = 318800 \): \[ 6400x - 18x^2 - 400 = 318800 \] Rearranging the equation: \[ -18x^2 + 6400x - 319200 = 0 \] Divide the entire equation by -2 to simplify: \[ 9x^2 - 3200x + 159600 = 0 \] Using the quadratic formula: \[ x = \frac{3200 \pm \sqrt{3200^2 - 4 \cdot 9 \cdot 159600}}{2 \cdot 9} \] Calculate the discriminant: \[ D = 3200^2 - 4 \cdot 9 \cdot 159600 = 10,240,000 - 5,745,600 = 4,494,400 \] \[ \sqrt{D} = 2120 \] Thus, the solutions are: \[ x = \frac{3200 \pm 2120}{18} \] Which gives: \[ x_1 = \frac{5320}{18} \approx 295.56 \] \[ x_2 = \frac{1080}{18} = 60.00 \] **Answer for part (a):** \[ 60.00, 295.56 \] ### Part (b): Can a Profit of More Than \$318,800 Be Made? The profit function is a downward-opening parabola (since the coefficient of \( x^2 \) is negative). The maximum profit occurs at the vertex of the parabola. The vertex \( x \) value is: \[ x = \frac{-b}{2a} = \frac{-6400}{2 \cdot (-18)} = \frac{6400}{36} \approx 177.78 \] Plugging \( x \approx 177.78 \) back into the profit function to find the maximum profit: \[ P_{\text{max}} = 6400(177.78) - 18(177.78)^2 - 400 \approx 568,488.89 \] Since the maximum profit (\$568,488.89) is greater than \$318,800: **Answer for part (b):** \[ \text{Yes} \] ### Final Answer **(a)** 60.00, 295.56 **(b)** Yes