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

Ask by Ramos Pope. in the United States

Mar 11,2025

**(A)** The average cost per vest is obtained by dividing the total cost by the number of vests, \( x \). That is, \[ \overline{C}(x) = \frac{C(x)}{x} = \frac{0.05x^2 + 10x + 405}{x}. \] This can be simplified by dividing each term in the numerator by \( x \): \[ \overline{C}(x) = 0.05x + 10 + \frac{405}{x}. \] **(B)** 1. **Finding the critical numbers:** To find the critical numbers of \(\overline{C}(x)\), we first compute its derivative. \[ \overline{C}(x) = 0.05x + 10 + \frac{405}{x}. \] Write \(\frac{405}{x}\) as \(405x^{-1}\) and differentiate term by term: \[ \overline{C}'(x) = 0.05 + 0 - 405x^{-2} = 0.05 - \frac{405}{x^2}. \] Set the derivative equal to zero: \[ 0.05 - \frac{405}{x^2} = 0. \] Solve for \( x \): \[ 0.05 = \frac{405}{x^2} \quad \Longrightarrow \quad x^2 = \frac{405}{0.05}. \] Calculate: \[ x^2 = 405 \times 20 = 8100 \quad \Longrightarrow \quad x = \sqrt{8100} = 90. \] Since \( x > 0 \) and \( x < 200 \), the only critical number is \[ x = 90. \] 2. **Intervals of decreasing and increasing behavior:** Analyze the sign of \(\overline{C}'(x)\): - For \( x \) in the interval \( (0, 90) \): Choose a test value, say \( x = 50 \): \[ \overline{C}'(50) = 0.05 - \frac{405}{50^2} = 0.05 - \frac{405}{2500} = 0.05 - 0.162 = -0.112. \] Since \(\overline{C}'(50) < 0\), the average cost function is **decreasing** on \( (0, 90) \). - For \( x \) in the interval \( (90, 200) \): Choose a test value, say \( x = 100 \): \[ \overline{C}'(100) = 0.05 - \frac{405}{100^2} = 0.05 - \frac{405}{10000} = 0.05 - 0.0405 = 0.0095. \] Since \(\overline{C}'(100) > 0\), the average cost function is **increasing** on \( (90, 200) \). 3. **Local extrema:** Since \(\overline{C}(x)\) is decreasing on \( (0, 90) \) and increasing on \( (90, 200) \), the critical number \( x = 90 \) corresponds to a local minimum. **Final Answers:** - (A) \[ \overline{C}(x) = 0.05x + 10 + \frac{405}{x}. \] - (B) - The critical number is \( x = 90 \). - The average cost per vest is decreasing on the interval \( (0, 90) \). - The average cost per vest is increasing on the interval \( (90, 200) \). - There is a local minimum at \( x = 90 \).