The problem provides the cost function $c(x) = 6x^3 - 36x^2 + 20,000x$, and we are asked to find the production level $x$ that minimizes the average cost and the value of the average cost at this level.

Steps to solve:

1. Find the average cost function: The average cost function $AC(x)$ is given by: $AC(x) = \frac{c(x)}{x} = \frac{6x^3 - 36x^2 + 20,000x}{x}$ Simplifying: $AC(x) = 6x^2 - 36x + 20,000$

2. Minimize the average cost: To minimize $AC(x)$, we first take its derivative and set it equal to zero: $AC'(x) = \frac{d}{dx}(6x^2 - 36x + 20,000) = 12x - 36$ Set $AC'(x) = 0$ to find the critical points: $12x - 36 = 0$ $x = 3$

3. Determine if this is a minimum: To confirm whether this critical point is a minimum, we check the second derivative of $AC(x)$: $AC''(x) = \frac{d}{dx}(12x - 36) = 12$ Since $AC''(x) = 12 > 0$, the function is concave up, confirming that $x = 3$ is a minimum.

4. Find the average cost at this level: Substitute $x = 3$ into the average cost function: $AC(3) = 6(3)^2 - 36(3) + 20,000$ $AC(3) = 6(9) - 108 + 20,000 = 54 - 108 + 20,000 = 19,946$

Final answers:

• The production level that minimizes the average cost is $x = 3$.
• The average cost at this production level is $19,946$.

Do you need further details on any step or have any questions?

Here are 5 related questions:

1. How do we find the second derivative of a function?
2. What does it mean if the second derivative is positive at a critical point?
3. How is the average cost function derived from the total cost function?
4. Why is setting the derivative of the average cost equal to zero important in optimization?
5. Can you apply this method to different forms of cost functions?

Tip: Always remember to check the second derivative when optimizing to confirm whether you have found a maximum or minimum.