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The problem in the image provides the following function for the average cost \( C(x) \) of producing \( x \) riding lawn mowers:

\[
C(x) = 0.8x^2 + 27x - 248 + \frac{2400}{x}
\]

### Part (a): Minimizing the Average Cost
To minimize the average cost, we need to find the value of \( x \) that minimizes \( C(x) \). This involves finding the derivative of \( C(x) \) with respect to \( x \), setting it equal to zero, and solving for \( x \).

### Part (b): Minimum Average Cost
Once we find the value of \( x \) that minimizes the average cost, we can substitute this value back into the original function \( C(x) \) to find the minimum average cost.

Let me perform these calculations for you.

The average cost per hour in dollars of producing x riding lawn mowers is given by the function C(x) = 0.8x^2 + 27x - 248 + 2400/x. Use a graphing utility to determine the number of riding lawn mowers to produce in order to minimize the average cost. What is the minimum average cost?

Learn how to use calculus to minimize the average cost function C(x) = 0.8x^2 + 27x - 248 + 2400/x for producing riding lawn mowers. This problem involves key concepts in algebra and calculus, including derivative tests for optimization. Suitable for advanced high school or college-level calculus students.

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