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

Ask by Maxwell Wood. in the United States

Mar 11,2025

To find the average cost per vest, we first calculate the average cost function \( \overline{C}(x) \). The total cost \( C(x) = 0.05x^2 + 10x + 405 \). The average cost is given by: \[ \overline{C}(x) = \frac{C(x)}{x} = \frac{0.05x^2 + 10x + 405}{x} = 0.05x + 10 + \frac{405}{x} \] Now, we can tackle the problem step by step. ### (A) Average Cost per Vest The average cost function: \[ \overline{C}(x) = 0.05x + 10 + \frac{405}{x} \] is defined for \( 0 < x < 200 \). ### (B) Critical Numbers and Intervals To find the critical numbers, we differentiate \( \overline{C}(x) \): \[ \overline{C}'(x) = 0.05 - \frac{405}{x^2} \] Setting the derivative equal to zero to find critical points: \[ 0.05 - \frac{405}{x^2} = 0 \implies 0.05x^2 = 405 \implies x^2 = \frac{405}{0.05} = 8100 \implies x = 90 \] The critical number is \( x = 90 \). Next, we determine the intervals of increase and decrease using the first derivative test. To do this, let's evaluate \( \overline{C}'(x) \): - For \( x < 90 \): Choose \( x = 1 \): \[ \overline{C}'(1) = 0.05 - \frac{405}{1} = 0.05 - 405 < 0 \quad \text{(decreasing)} \] - For \( x > 90 \): Choose \( x = 100 \): \[ \overline{C}'(100) = 0.05 - \frac{405}{10000} > 0 \quad \text{(increasing)} \] Thus, we can conclude: - The average cost per vest is decreasing on the interval \( (0, 90) \). - The average cost per vest is increasing on the interval \( (90, 200) \). ### Local Extrema From our findings: - At \( x = 90 \), test for local extrema: \[ \overline{C}(90) = 0.05(90) + 10 + \frac{405}{90} = 4.5 + 10 + 4.5 = 19 \] Since \( \overline{C}(x) \) decreases before \( x = 90 \) and increases after, we identify \( x = 90 \) as a local minimum. Ultimately, the answer choices are: - **The average cost per vest is decreasing on the interval(s)**: \( (0, 90) \) - **The average cost per vest is increasing on the interval(s)**: \( (90, 200) \) - **Find the local extrema**: B. \( \overline{C}(90) = 19 \) is a local minimum. There is no local maximum.