The average cost function is

¯C(q)=q3−28q2+202qq=q2−28q+202C¯(q)=q3−28q2+202qq=q2−28q+202

We differentiate this function and set it equal to 0 to find the production level that minimizes the average cost.

ddq(q2−28q+202)=02q−28=02q=28q=14 unitsddq(q2−28q+202)=02q−28=02q=28q=14 units

which is option d.

Then the minimum cost is

C(14)=142−28(14)+202=$6C(14)=142−28(14)+202=$6

per item, which is option b.

Average Cost:

The total cost of producing qq items is given by the cost function C(q)C(q). Since the average cost is the total cost divided by the number of items produced, it is given by

¯C(q)=C(q)qC¯(q)=C(q)q

To minimize/maximize any function, we can use the usual differentiation technique of finding where the derivative equals 0.