3.Given a total cost function: C= Q3 - 3Q2 + 8Q + 48 ,then find, a.TFC, TVC ,

Question

b.AFC,AVC, AC and MC
c.the minimum point of MC and the minimum point of AVC
Determine the level of output for which MC=AVC.

Answer 1

a. To find TFC (Total Fixed Cost), we need to determine the cost at zero output. From the given cost function, we can find C(0) as follows:

C(0) = (0^3) - 3(0^2) + 8(0) + 48
C(0) = 0 - 3(0) + 0 + 48
C(0) = 48

Therefore, TFC = 48.

To find TVC (Total Variable Cost), we subtract TFC from the total cost. So, TVC = C - TFC.

TVC = Q^3 - 3Q^2 + 8Q + 48 - 48
TVC = Q^3 - 3Q^2 + 8Q

b. AFC (Average Fixed Cost) is calculated by dividing TFC by the output quantity (Q). So, AFC = TFC / Q.

AFC = 48 / Q

AVC (Average Variable Cost) is calculated by dividing TVC by the output quantity (Q). So, AVC = TVC / Q.

AVC = (Q^3 - 3Q^2 + 8Q) / Q
AVC = Q^2 - 3Q + 8

AC (Average Cost) is calculated by dividing TC (Total Cost) by the output quantity (Q). So, AC = C / Q.

AC = (Q^3 - 3Q^2 + 8Q + 48) / Q
AC = Q^2 - 3Q + 8 + 48/Q

MC (Marginal Cost) is calculated by taking the derivative of the total cost function with respect to the output quantity (Q). So, MC = dC/dQ.

MC = 3Q^2 - 6Q + 8

c. To find the minimum point of MC, we need to find the Q value where MC = 0.
3Q^2 - 6Q + 8 = 0

Using the quadratic formula, we can solve for Q.
Q = (-b ± √(b^2 - 4ac)) / (2a)

Q = (-(-6) ± √((-6)^2 - 4(3)(8))) / (2(3))
Q = (6 ± √(36 - 96)) / 6
Q = (6 ± √(-60)) / 6

Since the square root of a negative number is not real, there are no real solutions for Q. Thus, there is no minimum point of MC.

To find the minimum point of AVC, we need to find the Q value where AVC is at its minimum. We can find this by taking the derivative of AVC (dAVC/dQ) and finding the Q value where it equals zero.

dAVC/dQ = 2Q - 3

Setting this equal to zero and solving for Q:
2Q - 3 = 0
2Q = 3
Q = 3/2

Therefore, the minimum point of AVC is at Q = 3/2.

d. To find the level of output for which MC = AVC, set the two equations equal to each other and solve for Q:

3Q^2 - 6Q + 8 = Q^2 - 3Q + 8

Subtract Q^2 and 8 from both sides:
2Q^2 - 3Q = 0

Factor out Q:
Q(2Q - 3) = 0

Since Q = 0 is not a valid solution, we solve for Q:
2Q - 3 = 0
2Q = 3
Q = 3/2

Therefore, the level of output for which MC = AVC is Q = 3/2.