Question

suppose that ( c(x)=3x^{3}-18x^{2}+14,000x ) is the cost of manufacturing ( x ) items. find a production level that will minimize the average cost of making ( x ) items. the production level that minimizes the average cost of making ( x ) items is ( x=) (simplify your answer.)

Explanation:

Step1: Find the average cost function

The average cost function \( \overline{c}(x)=\frac{c(x)}{x} \). Given \( c(x) = 3x^{3}-18x^{2}+14000x \), then \( \overline{c}(x)=\frac{3x^{3}-18x^{2}+14000x}{x}=3x^{2}-18x + 14000\).

Step2: Differentiate the average cost function

Differentiate \( \overline{c}(x) \) with respect to \( x \). Using the power rule \( (x^{n})^\prime=nx^{n - 1} \), we have \( \overline{c}^\prime(x)=(3x^{2}-18x + 14000)^\prime=6x-18\).

Step3: Find the critical points

Set \( \overline{c}^\prime(x)=0 \), so \( 6x-18 = 0 \). Solving for \( x \):

$$
LATEXBLOCK0
$$

Step4: Check the second - derivative

Differentiate \( \overline{c}^\prime(x) \) to get the second - derivative. \( \overline{c}^{\prime\prime}(x)=(6x - 18)^\prime=6>0 \). Since \( \overline{c}^{\prime\prime}(3)=6>0 \), the function \( \overline{c}(x) \) has a minimum at \( x = 3 \).

Answer:

\(3\)