- The cost function is defined as \(f(x) = \frac{4}{3}x^3 - \frac{15}{2}x^2 + \frac{353}{6}x + 57\).
- The objective is to find \(x\) such that \(f(x) = 310\) (representing \(\$310,000\)).
- By calculating the cost for each option, we find that \(f(4.75) \approx 310.14\), which is the closest value to 310.
- Therefore, the operation time required is \(\boxed{4.75}\).
Explanation
- Problem Analysis
We are given the function \(f(x) = \frac{4}{3}x^3 - \frac{15}{2}x^2 + \frac{353}{6}x + 57\), which models the cost of operating a factory for \(x\) hours, where the cost is measured in thousands of dollars. We aim to find the value of \(x\) such that the cost is \(\$310,000\). Since the cost function is in thousands of dollars, we need to set \(f(x) = 310\) and solve for \(x\). The given table confirms the function values at small integer hours, for instance at \(x=1\), \(f(1) = \frac{4}{3} - \frac{7.5} + \frac{353}{6} + 57 \approx 110\), which is consistent with the cost of 110 (thousand dollars).
- Evaluating and Comparing Results
We need to determine the value of \(x\) from the provided options (4.25, 4.5, 4.75, 5.0) that makes the cost \(f(x)\) approximately \(310\). We evaluate the function \(f(x)\) for each given option:
\(f(4.25) = \frac{4}{3}(4.25)^3 - 7.5(4.25)^2 + \frac{353}{6}(4.25) + 57 \approx 273.93\)
\(f(4.5) = \frac{4}{3}(4.5)^3 - 7.5(4.5)^2 + \frac{353}{6}(4.5) + 57 = 291.375\)
\(f(4.75) = \frac{4}{3}(4.75)^3 - 7.5(4.75)^2 + \frac{353}{6}(4.75) + 57 \approx 310.14\)
\(f(5.0) = \frac{4}{3}(5.0)^3 - 7.5(5.0)^2 + \frac{353}{6}(5.0) + 57 \approx 330.33\)
Comparing these values to our target cost of 310, we observe that \(f(4.75) \approx 310.14\) is the closest to 310.
- Conclusion
Based on the calculations, the number of hours of operation that results in a cost of approximately \(\$310,000\) is 4.75. We have evaluated the function for the provided options and identified that 4.75 is the value that yields a cost closest to the target. Thus, the final answer is \(\boxed{4.75}\).
Examples
This type of cost analysis is fundamental in industrial engineering and business management. For example, a delivery company might model the cost of fuel based on the time spent driving. By finding the number of hours that keeps costs within a specific budget, managers can optimize delivery schedules to ensure profitability. Such algebraic models help in making informed, data-driven decisions every day.