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The given marginal cost function is:

\[
C(x) = 3x^2 - 140x + 2900
\]

To minimize the marginal cost, we need to find the critical points by taking the derivative of the function and setting it equal to zero.

1. Take the derivative of \(C(x)\) with respect to \(x\):
   \[
   C'(x) = \frac{d}{dx}(3x^2 - 140x + 2900)
   \]

2. Set \(C'(x) = 0\) to find the critical points.

Let me calculate this.The critical point is \( x = \frac{70}{3} \approx 23.33 \).

Therefore, the company should produce approximately 23 or 24 units to minimize the marginal cost.

Would you like further details or clarification?

Here are 5 related questions you could explore:
1. What is the value of the marginal cost at the minimum point?
2. How does the second derivative test confirm that this is a minimum?
3. What happens to the marginal cost if the company produces more than the optimal number of units?
4. How is the behavior of this quadratic cost function related to real-world production costs?
5. What does the constant term in the cost function represent?

**Tip:** When minimizing cost functions, always check the second derivative to confirm whether the critical point is a minimum or maximum.

The marginal cost to produce one additional (generic consumer good) after having produced x of them already is modeled by C(x) = 3x^2 - 140x + 2900. How many (generic consumer good)s should a company produce to minimize the marginal cost?

This problem demonstrates how to minimize the marginal cost function C(x) = 3x^2 - 140x + 2900 using calculus concepts like derivatives and critical points. Learn the detailed step-by-step solution, including how to find and interpret the critical point.

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