MC = 125 - 0.42Q + 0.0021Q^2

Given the estimated marginal cost function, what are the total, variable, and average cost functions?

At what output level does AVC reach its minimum value in the long run? What is the value of AVC at its minimum point?

MC is the first derivitive of the total cost function. So, pretend you are integrating.

TC = 125Q + (.42/2)Q^2 + (.0021/3)Q^3

Divide by Q to get AVC.

Use standard calculus to find the minima of the AVC function.

To find the total cost function, we need to integrate the marginal cost function with respect to output (Q). Integrating the given equation will give us the total cost function (TC).

TC = ∫ (MC dQ)
TC = ∫ (125 - 0.42Q + 0.0021Q^2) dQ

To find the integral, we can apply the power rule of integration.

TC = 125Q - 0.42(Q^2)/2 + 0.0021(Q^3)/3 + C

Where C is the constant of integration.

The variable cost function (VC) is the cost incurred as a result of producing an additional unit. To find VC, we need to subtract the fixed cost (FC) from the total cost (TC).

VC = TC - FC

The average cost function (AC) is the cost per unit of output. To find AC, we need to divide the total cost (TC) by the quantity produced (Q).

AC = TC / Q

To find the output level at which AVC reaches its minimum value in the long run, we need to differentiate the average variable cost (AVC) function with respect to Q and set it equal to zero. The result will give us the minimum point of AVC.

Let's calculate AVC using the formulas:

AVC = VC / Q
AVC = (TC - FC) / Q

Now, let's differentiate AVC with respect to Q and solve for Q when the derivative equals zero:

dAVC / dQ = [d/dQ (TC - FC)] / Q = 0

To find out the value of AVC at its minimum point, substitute the value of Q into the average variable cost function:

AVC = ((125Q - 0.42(Q^2)/2 + 0.0021(Q^3)/3 + C) - FC) / Q

Please provide the value of FC, and I can calculate the remaining values for you.