The marginal cost of a product can be thought of as the cost of producing one additional unit of output.  For​ example, if the marginal cost of producing the 50th product is​ $6.20, it cost​ $6.20 to increase production from 49 to 50 units of output.  Suppose the marginal cost C​ (in dollars) to produce x thousand mp3 players is given by the function Upper C left parenthesis x right parenthesis equals x squared minus 140 x plus 7000.
A.  How many players should be produced to minimize the marginal​ cost?
B.  What is the minimum marginal​ cost?

Let's analyze and solve the problem step by step.

We are given the marginal cost function:

\[
C(x) = x^2 - 140x + 7000
\]

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

To find the number of players $x$ that minimize the marginal cost, we need to find the critical points of the function by taking its derivative and setting it equal to zero.

1. **First, take the derivative of the marginal cost function:**

\[
C'(x) = \frac{d}{dx}(x^2 - 140x + 7000)
\]

\[
C'(x) = 2x - 140
\]

2. **Set the derivative equal to zero to find the critical points:**

\[
2x - 140 = 0
\]

Solving for $x$:

\[
2x = 140
\]

\[
x = 70
\]

So, the marginal cost is minimized when $x = 70$, meaning 70 thousand MP3 players should be produced to minimize the marginal cost.

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

To find the minimum marginal cost, substitute $x = 70$ back into the original marginal cost function:

\[
C(70) = (70)^2 - 140(70) + 7000
\]

Calculating this:

\[
C(70) = 4900 - 9800 + 7000
\]

\[
C(70) = 2100
\]

Thus, the minimum marginal cost is **$2100**.

### Summary:
- **A.** 70 thousand players should be produced to minimize the marginal cost.
- **B.** The minimum marginal cost is **$2100**.

Would you like more details or further explanation?

Here are 5 related questions to expand your understanding:
1. What is the interpretation of a critical point in optimization problems?
2. How do we know the critical point at $x = 70$ is a minimum, not a maximum?
3. How would changes in the marginal cost function affect the optimal production level?
4. What role does the second derivative play in determining minima and maxima?
5. How can we use the marginal cost function to determine the cost for producing a specific number of units?

**Tip:** The second derivative test helps confirm if a critical point is a minimum or maximum by checking the concavity of the function at that point.

The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand mp3 players is given by the function C(x) = x^2 - 140x + 7000. A. How many players should be produced to minimize the marginal cost? B. What is the minimum marginal cost?

Explore how to minimize the marginal cost of producing mp3 players using the quadratic cost function C(x) = x^2 - 140x + 7000. Learn how to use derivatives to find the critical point and calculate the minimum marginal cost. Suitable for advanced high school students (Grades 11-12).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation