Brewer Griffiths
03/10/2023 · Middle School
Suppose that \( c(x)=3 x^{3}-18 x^{2}+24,000 x \) is the cost of manufacturing \( x \) items. Find a production level that will minimize the average cost of making \( x \) items. At this level, the average cost of making \( x \) items is \( \$ \) (Simplify your answer.)
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Step-by-step Solution
To find the production level that minimizes the average cost of making \( x \) items, we first need to determine the average cost function. The average cost \( A(x) \) is given by the total cost \( c(x) \) divided by the number of items \( x \):
\[
A(x) = \frac{c(x)}{x} = \frac{3x^3 - 18x^2 + 24000x}{x}
\]
Simplifying this, we get:
\[
A(x) = 3x^2 - 18x + 24000
\]
Next, we need to find the critical points of \( A(x) \) by taking the derivative and setting it to zero:
\[
A'(x) = \frac{d}{dx}(3x^2 - 18x + 24000) = 6x - 18
\]
Setting the derivative equal to zero to find critical points:
\[
6x - 18 = 0
\]
Solving for \( x \):
\[
6x = 18 \implies x = 3
\]
To confirm that this critical point is a minimum, we can check the second derivative:
\[
A''(x) = \frac{d^2}{dx^2}(3x^2 - 18x + 24000) = 6
\]
Since \( A''(x) = 6 > 0 \), the function \( A(x) \) is concave up at \( x = 3 \), indicating that this point is indeed a minimum.
Now, we can find the average cost at this production level:
\[
A(3) = 3(3^2) - 18(3) + 24000
\]
Calculating each term:
\[
A(3) = 3(9) - 54 + 24000 = 27 - 54 + 24000 = -27 + 24000 = 23973
\]
Thus, the average cost of making \( x \) items at the production level that minimizes the average cost is:
\[
\boxed{23973}
\]
Quick Answer
The average cost of making \( x \) items at the production level that minimizes the average cost is \( \$23973 \).
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