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The problem states that the unit cost \( C(x) \), in dollars, to make each copy machine is given by the quadratic function:

\[
C(x) = x^2 - 580x + 93,767
\]

where \( x \) is the number of machines made. We need to find the number of machines, \( x \), that minimizes the unit cost.

### Solution

Since \( C(x) \) is a quadratic function with a positive \( x^2 \) coefficient, it has a minimum value at its vertex. For a quadratic function in the form \( ax^2 + bx + c \), the \( x \)-coordinate of the vertex (which gives the minimum or maximum point) is:

\[
x = -\frac{b}{2a}
\]

In this case:
- \( a = 1 \)
- \( b = -580 \)

Substitute these values into the formula:

\[
x = -\frac{-580}{2 \cdot 1} = \frac{580}{2} = 290
\]

### Answer
The minimum unit cost is achieved when \( x = 290 \) machines are made.

Let me know if you would like more details, or have any questions.

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### Related Questions
1. What would the minimum unit cost be when \( x = 290 \) machines are produced?
2. How does the unit cost change as more machines are produced beyond the minimum cost point?
3. If the production cost function changes, how would you determine the new minimum number of machines?
4. Can the function \( C(x) \) have a maximum value within a specific range of \( x \)?
5. How does changing the coefficient of \( x \) in \( C(x) \) affect the minimum cost point?

### Tip
For any quadratic function \( ax^2 + bx + c \), the vertex formula \( x = -\frac{b}{2a} \) is useful to find the minimum or maximum point, depending on the direction of the parabola.

A supply company manufactures copy machines. The unit cost C (the cost in dollars to make each copy machine) depends on the number of machines made. If x machines are made, then the unit cost is given by the function C(x) = x^2 - 580x + 93,767. How many machines must be made to minimize the unit cost?

Discover how to find the number of machines needed to minimize the unit cost in manufacturing with the function C(x) = x^2 - 580x + 93,767. Learn the application of quadratic functions and optimization techniques for Grades 9-12.

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