Question

Total cost C(x) of a firm is
C(Q)-Q3-12Q2+60Q+1200 Where Q denotes the output

1.   AC and Slope of AC
2.   MC and the Minimum of MC
3.   Value of Q which MC – AVC where VC denotes the variable cost
4.   Show that MC cuts minimum point of the AVC

Answer 1

Ans. Cost function, C = Q^3 - 12Q^2 + 60Q + 1200

a) Average cost, AC = C/Q = Q^2 - 12*Q + 60 + 1200/Q

Slope of AC =  dAC/dQ = 2Q - 12 - 1200/Q^2

b) Marginal Cost, MC = dC/dQ = 3Q^2 - 24Q + 60

For minimum point of MC,

dMC/dQ = 6Q - 24 = 0

=> Q = 4 units

Thus, MC is minimum at output level of 4 units

c) VC = C - C (at Q = 0) = Q^3 - 12Q^2 + 60Q

=> AVC = VC/Q = Q^2 - 12Q + 60

At MC = AVC

=> 3Q^2 - 24Q + 60 = Q^2 - 12Q + 60

=> 2Q^2 -12Q = 0

=> Q = 6 units

d) For minimum point of AVC,

dAVC/dQ = 2Q - 12 = 0

=> Q = 6 units

Thus, AVC is minimum at 6 units of output and MC cuts AVC at 6 units.