A firm's fixed costs are 4545 and the variable costs are 3Q3Q per unit

a) Since we're told the fixed and variable costs for this firm, we can construct the total cost function by adding them together. Note that we have to multiply the variable cost by the quantity, as the wording of the problem states that they are "3Q per unit", and each unit is 1Q.

TC=45+(3Q)Q=45+3Q2TC=45+(3Q)Q=45+3Q2

Both the marginal cost and the average cost functions are defined using this function. First, the marginal cost function is its derivative.

MC=6QMC=6Q

Second, the average cost function is defined as the quotient of the total cost over the quantity.

AC=45+3Q2Q=45Q+3QAC=45+3Q2Q=45Q+3Q

b) We can minimize the average cost function using the first derivative. First, let's differentiate this function.

dACdQ=−45Q2+3dACdQ=−45Q2+3

Setting this equal to zero will yield the critical points of the function. Since the quantity cannot be zero, we can solve for the critical points as follows.

−45Q2+33=45Q23Q2=45Q2=15Q=±√15−45Q2+33=45Q23Q2=45Q2=15Q=±15

We can ignore the negative result, since we can't produce a negative quantity, either. Thus, we only have one critical point to consider. The Second Derivative Test can confirm or deny whether or not this is a minimum of the average cost function.

d2ACdQ2=90Q3d2ACdQ2(√15)=90(√15)3≈1.594d2ACdQ2=90Q3d2ACdQ2(15)=90(15)3≈1.594

Since this is a positive value, this is the quantity that minimizes the average cost.

Therefore, the quantity that minimizes the average cost is Q=√15Q=15.