I am so lost for this homework question, can anyone give me some help, thanks. T
ID: 1209212 • Letter: I
Question
I am so lost for this homework question, can anyone give me some help, thanks.
The average cost per item to produce q items is given by a(q) = 0.01q^2 0.6q + 13, for q > 0.
(a) What is the total cost, C(q), of producing q goods?
(b) What is the minimum marginal cost? What is the practical interpretation of this result?
(c) At what production level is the average cost a minimum? What is the lowest average cost?
(d) Compute the marginal cost at q = 30. How does this relate to your answer to part (c)? Explain this relationship both analytically and in words.
Explanation / Answer
The average cost per item to produce q items is given by a(q) = 0.01q^2 0.6q + 13, for q > 0.
(a) Average total cost is the total cost divided by the number of units produced. This implies AC = TC/q. To get TC from AC, multiply AC by q.
The total cost, C(q), of producing q goods = AC*q
C(q) = a(q)*q
=(0.01q2 0.6q + 13)q, for q > 0
=0.01q3 0.6q2 + 13q, for q > 0.
Hence total cost is c(q) = 0.01q3 0.6q2 + 13q, for q > 0.
(b) Marginal Cost: It is the first derivative of the total cost with respect to the output. So Marginal cost is:
MC(q) = dc(q)/q
=d(0.01q3 0.6q2 + 13q)/q
=0.03q2 1.2q + 13
The minimum level of marginal cost is achieved when the derivative of marginal cost is set equal to zero.
MC'(q) = 0
0.06q - 1.2 = 0
q = 20
At q = 20, MC(20) = 0.03(20)2 1.2(20) + 13 = 25 - 24 = 1
Hence the minimum of marginal cost is achieved when q = 20 and minimum MC is $1
(c) Average cost is at its minimum when the derivative of Average cost is set equal to zero
AC'(q) = 0
0.02q 0.6 = 0
q = 30. At q = 30, AC(30) = 0.01(30)2 0.6*30 + 13 = 4
Hence the lowest average cost is $4 at q = 30
(d) The value of marginal cost at q = 30 is given by:
MC(30) = 0.03(30)2 1.2(30) + 13 = 4
Note that MC = AC when AC is at its minimum level. This provides the general economic result:
When MC>AC, AC rises
When MC<AC, AC falls
When MC = AC, AC is at its minimum.
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