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The marginal cost function is given as $ C(x) = x^2 - 160x + 7000 $, where $ x $ represents the number of thousand MP3 players produced.

### A. How many players should be produced to minimize the marginal cost?

To minimize the marginal cost, we need to find the value of $ x $ that minimizes the quadratic function. This can be done by finding the vertex of the parabola. For a quadratic equation $ C(x) = ax^2 + bx + c $, the x-coordinate of the vertex is given by:

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

Here, $ a = 1 $ and $ b = -160 $, so:

\[
x = \frac{-(-160)}{2(1)} = \frac{160}{2} = 80
\]

Thus, the marginal cost is minimized when 80 thousand MP3 players are produced.

### B. What is the minimum marginal cost?

To find the minimum marginal cost, substitute $ x = 80 $ into the cost function:

\[
C(80) = (80)^2 - 160(80) + 7000
\]
\[
C(80) = 6400 - 12800 + 7000 = 21200 - 12800 = 8400
\]

Thus, the minimum marginal cost is \$8400.

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Let me know if you need further details or clarifications. Here are some related questions for further exploration:

1. How does the shape of a quadratic function affect the marginal cost in production economics?
2. What are the implications of minimizing marginal cost in a real-world production scenario?
3. How would the minimum cost change if the production cost function included a cubic term?
4. How can we derive the marginal cost from total cost functions?
5. What is the significance of marginal cost in pricing strategies?

**Tip:** In a quadratic cost function, the minimum or maximum point is always at the vertex. Make sure to know the vertex formula for quick optimization!

Suppose the marginal cost C (in dollars) to produce x thousand MP3 players is given by the function C(x) = x^2 - 160x + 7000. How many players should be produced to minimize the marginal cost? What is the minimum marginal cost?

Learn how to minimize the marginal cost of producing MP3 players using the quadratic cost function C(x) = x^2 - 160x + 7000. This example demonstrates how to apply the vertex formula to optimize production. Suitable for high school students learning algebra and optimization techniques.

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