A manufacturer incurs the following costs in producing \( x \) water ski vests in one day, for \( 0

Ask by Lawson Parry. in the United States

Mar 12,2025

To find the average cost per vest, \( \overline{C}(x) \), we start with the total cost function \( C(x) = 0.05x^2 + 20x + 245 \). The average cost is then given by: \[ \overline{C}(x) = \frac{C(x)}{x} = \frac{0.05x^2 + 20x + 245}{x} = 0.05x + 20 + \frac{245}{x} \] Now, for part (B), to find critical numbers, let’s differentiate \( \overline{C}(x) \): \[ \overline{C}'(x) = 0.05 - \frac{245}{x^2} \] Setting the derivative to zero for critical points gives: \[ 0.05 - \frac{245}{x^2} = 0 \implies \frac{245}{x^2} = 0.05 \implies x^2 = \frac{245}{0.05} \implies x^2 = 4900 \implies x = 70 \] Thus, the critical number is \( x = 70 \). To determine intervals of increase and decrease, we can analyze the derivative: - For \( x < 70 \), \( \overline{C}'(x) > 0\) (increasing). - For \( x > 70 \), \( \overline{C}'(x) < 0\) (decreasing). This tells us that \( \overline{C}(x) \) has a local minimum at \( x = 70 \). Therefore, the average cost \( \overline{C}(70) = 0.05(70) + 20 + \frac{245}{70} = 3.5 + 20 + 3.5 = 27 \). Let's summarize the findings: A. \( \overline{C}(x) \) is decreasing on \( (70, 175) \), \( \overline{C}(x) \) is increasing on \( (0, 70) \), and the local minimum is at \( x = 70 \). B. \( \overline{C}(70) = 27 \). C. There are no local maxima as the function only possesses one local minimum.