Pregunta
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A manufacturer incurs the following costs in producing \( x \) water ski vests in one day, for \( 0
Ask by Olson Osborne. in the United States
Mar 12,2025

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**(A)** The average cost per vest is \( \overline{C}(x) = 0.05x + 20 + \frac{245}{x} \). **(B)** 1. **Critical Number:** \( x = 70 \) 2. **Intervals:** - Decreasing: \( 0 < x < 70 \) - Increasing: \( 70 < x < 175 \) 3. **Local Minimum:** At \( x = 70 \)

Solución

**(A)** The average cost per vest is given by \[ \overline{C}(x)=\frac{C(x)}{x}=\frac{0.05x^2+20x+245}{x}=0.05x+20+\frac{245}{x}. \] **(B)** 1. **Differentiate \(\overline{C}(x)\):** We have \[ \overline{C}(x)=0.05x+20+245x^{-1}. \] Differentiate term-by-term with respect to \(x\): \[ \overline{C}'(x)=0.05-245x^{-2} = 0.05-\frac{245}{x^2}. \] 2. **Find the critical numbers:** Set the derivative equal to 0: \[ 0.05-\frac{245}{x^2}=0. \] Solve for \(x\): \[ 0.05=\frac{245}{x^2} \quad\Longrightarrow\quad x^2=\frac{245}{0.05}. \] Simplify the fraction: \[ x^2=\frac{245}{0.05}=245\cdot 20=4900. \] Take the square root (considering \(x>0\)): \[ x=70. \] Thus, \(x=70\) is the only critical number in the domain \(00\), \(\overline{C}(x)\) is increasing on \((70,175)\). 4. **Local Extrema:** Because \(\overline{C}(x)\) decreases for \(x<70\) and increases for \(x>70\), the critical number \(x=70\) is a local minimum. **Summary:** - \(\overline{C}(x)=0.05x+20+\frac{245}{x}\) - Critical number: \(x=70\) - \(\overline{C}(x)\) is decreasing on the interval \((0,70)\) - \(\overline{C}(x)\) is increasing on the interval \((70,175)\) - There is a local minimum at \(x=70\).

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Bonus Knowledge

To find the average cost \( \bar{C}(x) \) per vest when \( x \) vests are produced in one day, we divide the total cost \( C(x) \) by the number of vests produced \( x \): \[ \bar{C}(x) = \frac{C(x)}{x} = \frac{0.05x^2 + 20x + 245}{x} = 0.05x + 20 + \frac{245}{x} \] Thus, the average cost per vest is given by: \[ \bar{C}(x) = 0.05x + 20 + \frac{245}{x} \] Now, moving on to the next part: To find the critical numbers of \( \bar{C}(x) \), we need to take the derivative of \( \bar{C}(x) \) and set it to zero: \[ \bar{C}'(x) = 0.05 - \frac{245}{x^2} \] Setting \( \bar{C}'(x) \) to zero gives us: \[ 0 = 0.05 - \frac{245}{x^2} \] \[ \frac{245}{x^2} = 0.05 \rightarrow x^2 = \frac{245}{0.05} = 4900 \rightarrow x = 70 \] Next, determine the intervals of increase and decrease by analyzing the sign of \( \bar{C}'(x) \): 1. For \( 0 < x < 70 \): \( \bar{C}'(x) > 0 \) (the function is increasing). 2. For \( x > 70 \): \( \bar{C}'(x) < 0 \) (the function is decreasing). Thus, the average cost per vest is increasing on the interval \( (0, 70) \) and decreasing on \( (70, 175) \). The local minimum occurs at \( x = 70 \), which is the critical number we found. In summary: (A) \( \overline{\mathrm{C}}(\mathrm{x})=0.05x + 20 + \frac{245}{x} \) (B) The critical number is \( x = 70 \), the average cost is decreasing for \( x > 70 \), increasing for \( 0 < x < 70 \), and the local extremum occurs at \( x = 70 \).

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