Question

1. Given the following cost function, determine the level of (nonzero) output at which the average cost function in minimized, and the average cost at that level.
   
   AC = 520 -160Q + 20Q²
   
   (10 points)

• An Average Cost is at a minimum when its derivative function (measuring its change) is zero.

Answer 1

Answer
average cost is minimum where MC=ATC or dAC/dq=0
by the first method
The total cost =AC*Q=520Q-160Q^2+20Q^3
MC is the marginal cost which is the change in total cost, and it is found by differentiation
dTC/dq=520-320Q+60Q^2=MC
MC=AC
520-320Q+60Q^2=520-160Q+20Q^2
-160Q+40Q^2=0
Q=4

from the second method

A function is minimum at first derivative equal to zero
dAC/dQ=-160+40Q=0
40Q=160
Q=4
the AC is minimum at Q=4
AC=520-160*4+20*4^2
=200

the average total cost is $200