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 50 th product is \( \$ 620 \), 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 \( C(x)=x^{2}-120 x+8400 \). A. How many players should be produced to minimize the marginal cost? B. What is the minimum marginal cost? A. To minimize the marginal cost, \( \square \) thousand mp3 players should be produced. B. The minimum marginal cost is \( \$ \square \)
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To find the number of players that should be produced to minimize the marginal cost, we first need to take the derivative of the marginal cost function \( C(x) = x^2 - 120x + 8400 \). Setting the derivative \( C'(x) = 2x - 120 \) equal to zero gives us \( 2x - 120 = 0 \), leading to \( x = 60 \). So, to minimize the marginal cost, **60 thousand mp3 players should be produced**. To find the minimum marginal cost, we substitute \( x = 60 \) back into the original cost function: \( C(60) = (60)^2 - 120(60) + 8400 \). This calculates to \( C(60) = 3600 - 7200 + 8400 = 3600 \). Hence, **the minimum marginal cost is \( \$ 3600 \)**.
