To determine when the Average Variable Cost (AVC) is at its minimum for the given Total Cost (TC) function:

Total Cost (TC) function:

$TC = 40 + 82q - 6q^2 + 0.2q^3$

1. Separate Variable and Fixed Costs:

   • Fixed Cost (FC): $40$
   • Variable Cost (VC): $82q - 6q^2 + 0.2q^3$

   Average Variable Cost (AVC) is: $AVC = \frac{VC}{q} = 82 - 6q + 0.2q^2$

2. Minimizing AVC: To find the minimum value of $AVC$, differentiate $AVC$ with respect to $q$ and set the derivative equal to zero: $\frac{d(AVC)}{dq} = \frac{d}{dq}(82 - 6q + 0.2q^2)$

   $\frac{d(AVC)}{dq} = -6 + 0.4q$

   Set $\frac{d(AVC)}{dq} = 0$: $-6 + 0.4q = 0$

   Solve for $q$: $q = \frac{6}{0.4} = 15$

3. Verify Minimum: Take the second derivative of $AVC$: $\frac{d^2(AVC)}{dq^2} = \frac{d}{dq}(-6 + 0.4q) = 0.4$ Since $\frac{d^2(AVC)}{dq^2} > 0$, $AVC$ is at a minimum when $q = 15$.

Conclusion:

The Average Variable Cost (AVC) is at its minimum when $q = 15$.

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Would you like a detailed explanation of the process or further assistance with related questions?

5 Related Questions:

1. How is the marginal cost derived from the given TC function?
2. What is the significance of minimizing AVC in production decisions?
3. How does fixed cost affect the Average Total Cost (ATC)?
4. What happens to AVC if production increases indefinitely?
5. How would a change in the cubic term ($0.2q^3$) impact the AVC curve?

Tip:

Always check the second derivative to confirm whether a critical point is a minimum or maximum.