Production functions and cost functions are “two sides of the same coin” – and m
ID: 2440223 • Letter: P
Question
Production functions and cost functions are “two sides of the same coin” – and managers need to understand their production processes and constraints, and the implications for costs, thoroughly to make appropriate production and pricing decisions.
A production function relates inputs like labor and capital to output, thus taking the form Q = f(L, K). Cost functions in turn relate output to cost: C = f(Q). The technology employed in the production process thus becomes the key determinant of the firm’s cost structure, in conjunction with the costs of inputs. To become better acquainted with the concepts, we will take a simple production concept, specify the production function, and derive from that function the cost structure of the firm, for both the short run and the long run. Then we’ll use this information, along with a demand function, to decide how much the firm should produce and the price it should charge. Production functions and the associated cost functions can, of course, get complicated – as can the math that describes them. But in this problem the math is quite simple; it’s the logic in the relationships we want to focus on.
Picture a company that makes shirts using three inputs: labor, sewing machines, and cloth. To make the example as simple as possible, yet still fairly realistic, we’ll suppose that the firm must use these inputs in “fixed proportions”: it takes one worker one hour with a sewing machine and three yards of cloth to make a shirt.
This “fixed proportions” or “fixed coefficients” production function indicates that the inputs are perfect complements in production (as opposed to substitutes – you can’t substitute labor for a yard of cloth, for example). This technology is also called a "Leontief" production function, after a famous economist who did a lot of work on production processes.
We can write this production function in mathematical terms. Specifically, let Q be the number of shirts, L be the number of labor hours, K (for capital) be the number of sewing machine hours, and T (for raw materials) be the number of yards of cloth you have available. Then the production function we described above can be written as
Q = min { L, K, T/3 }
The operator “min” stands for minimum, meaning that you can produce the lesser of the amounts of L, K, and T/3 that you have. Consider an example: suppose you have available for production 8000 worker hours, 6000 sewing machine hours, and 12,000 yards of cloth. How many shirts can you make? Mathematically, you could say Q = min { 8000, 6000, 12000/3 } or thus Q = min { 8000, 6000, 4000 }. The minimum of these is 4000, which is the number of shirts you could make. Cloth becomes the constraining factor or “bottleneck” in the production process; you’ve got more sewing machines and more workers than you need for the amount of cloth that you’ve got.
(Sometimes people are thrown off a little by the term T/3; since it takes 3 yards of cloth, they want to write 3T – but that’s not correct. Think of the 1/3 multiplying the T as being the marginal product of a yard of cloth: if you’ve got a yard of cloth you can make one third of a shirt, not 3 shirts!)
If you actually wanted to make 4000 shirts, of course, you wouldn’t hire 8000 worker hours and rent 6000 sewing machine hours – that’s clearly wasteful. You’d want to have 4000 worker hours, 4000 machine hours, and 12,000 yards of cloth on hand, and no more. In the long run when we can choose all of our inputs, that’s exactly how we’d do it: we’d decide how many shirts we were going to make, then hire that many worker hours, rent the same number of sewing machine hours (or buy the equivalent capacity), and order three times that number of yards of cloth. Mathematically we’d say we want L = Q, K = Q. and T = 3Q.
In the short run, however, we are (by definition) stuck with a fixed quantity of some input, typically our capital. Thus our decision making in the short run is constrained by our capacity limitations and we may not operate as efficiently as we can in the long run.
Now let’s answer some questions about the production process:
1a. Given this production function, how many shirts can you make with the following quantities of inputs?
L
K
T
Q
4000
3500
15,000
6000
6000
6000
3000
6000
9000
1b. Now show the efficient levels of L, K, and T needed to produce each amount of Q:
L
K
T
Q
1000
8000
20,000
1c. Here’s slightly tricky problem: How would you write the fixed proportions production function if it took one hour of labor, half an hour of sewing machine time, and two yards of cloth to make a shirt?
Q =
Now we turn our attention to the corresponding cost functions, beginning with a long run scenario. Cost functions express the cost of production, C, as a function of output, Q. We need to know what the inputs cost per unit, so assume that L costs $6 per hour, K costs $1 per hour, and T costs $2 per yard.
2a. Based on original production function, Q = min { L, K, T/3 }, and using the most efficient mix of inputs possible, answer the following questions:
How much does it cost to make 1 shirt?
TC(1) =
How much does it cost to make 2 shirts?
TC(2) =
How much does it cost to make 3 shirts?
TC(3) =
How much does it cost to make 1000 shirts?
TC(1000) =
How much does it cost to make Q shirts?
TC(Q) =
The last answer above is your long run total cost function, and it is written as an expression of Q. Because the equation is linear (i.e., this cost function is a straight line, not a curve), we can tell that this production function features “constant returns to scale” – you can double your output, for example, by doubling your inputs, which will double your costs. This feature is not always realistic, but some production processes are in fact constant returns to scale functions, and mathematically these are the easiest ones to work with.
2b. Average total cost is found by dividing total cost by output, or ATC = TC/Q. Write the equation for long run average total cost:
ATC =
2c. Marginal cost is the cost of making an additional shirt, or MC = ?TC/?Q, which means the change in total cost divided by the change in output. Conveniently for a constant returns to scale production function, marginal cost is constant and equal to average total cost. The cost of making each additional shirt is:
MC =
Now let’s use this information to make decisions about production and pricing. Suppose the inverse demand for these shirts is given by P = 30 - Q/1000, and we want to know how many shirts the firm should make, what inputs it will need, and what price the firm should charge to maximize profits.
First we can use the “short-cut rule” (using the inverse demand function, “same intercept, twice the slope”) to find marginal revenue:
MR =
Setting MR = MC enables us to find the optimal quantity, and then the profit maximizing price:
Q* =
P* =
And since we know the production function, we can work back to find the optimal input levels for profit maximization:
L* =
K* =
T* =
The firm’s maximum profits in the long run are thus:
?* =
Now let’s pause for a little more nuanced question. You’ll recall from chapter 1 that economists think of profits differently from accountants, because economists think differently about costs. For economic (i.e., business) decision making, it’s important to take into account all opportunity costs, both explicit (i.e., accounting costs) and implicit costs. Sometimes this is difficult. Consider the following scenario:
Suppose the sewing machine has already been bought and paid for at a price of $1000, but that if the firm wasn't using it, it could be rented out at a price of $5/hr. Is it more appropriate to treat the sewing machine as a fixed cost at $1000, or as a variable cost at $5/hr? Why? How could this affect your production decision?
Earlier we solved the firm’s long run production problem. Now let’s think about how the situation might be different in the short run, in a couple of different scenarios. Let’s go back to the original set of conditions.
Demand: P = 30 - Q/1000
Production function: Q = min { L, K, T/3 }
Costs of inputs: Hourly wage rate for labor = $6
Hourly rental rate for sewing machines = $1
Cost of a yard of cloth = $2
Suppose that in Short Run Scenario 1, the firm has contractually obligated itself to pay for 5000 hours of sewing machine usage. Thus K is fixed at 5000 in the short run. (To be clear: this is a maximum amount of K that can be used in the short run; all 5000 hours must be paid for, but not all 5000 hours must actually be used.)
What will now be the firm’s total fixed cost of production in the short run (that is, the total obligation for the payment of the sewing machines?)
TFC =
And what will be the marginal cost of making each additional shirt, up to 5000 shirts? Since the sewing machines are now paid for, you only need to consider the labor and cloth costs per shirt.
MC =
Your total variable cost will be your marginal cost, which in this simple problem is constant, times the quantity produced. What is your total variable cost of producing up to 5000 shirts?
TVC =
Your total short run cost of production will be TFC + TVC, so write the expression for total cost below:
TC =
Keep in mind however, that this is only true up to your capacity constraint, which is 5000 shirts, because you only have 5000 sewing machine hours available. This is a constrained optimization problem.
In this setting, Short Run Scenario 1, how many shirts should (can) you produce, what price should you charge, how much of each input should you employ, and how much profit can you earn?
Q* =
P* =
L* =
K* =
T* =
?* =
And finally, let’s consider Short Run Scenario 2: you have contracted for (and are obligated to pay for) 10,000 hours of sewing machine time. Again, K = 10,000 represents a maximum but you don’t have to make 10,000 shirts just because you’ve already paid for the hours, if you don’t want to. Other conditions remain the same.
What will your cost structure look like now?
TFC =
TVC =
TC =
MC =
And once again using the MR = MC relationship to maximize profits, find your optimal level of output, price for shirts, input levels, and profit:
Q* =
P* =
L* =
K* =
T* =
?* =
Note that managing in the short run is, in a sense, easier – because there are fewer decisions to make, since some inputs are fixed. But managing in the long run leads to greater profits (as long as demand and input costs remain the same) because you have more flexibility with your inputs and can thus get just the right mix of resources.
L
K
T
Q
4000
3500
15,000
6000
6000
6000
3000
6000
9000
Explanation / Answer
1.a. Given that the production function is Q = min(L, K, T/3)
min(4000, 3500, 15000/3) = min(4000, 3500, 5000) = 3500.
Thus, Q = 3500
min(6000, 6000, 6000/3) = min(6000, 6000, 2000) = 2000
Thus, Q = 2000
min(3000, 6000, 9000/3) = min(3000, 6000, 3000) = 3000
Thus, Q = 3000
1.b. The efficient amounts of input required for the production of shirts can be mathematically expressed as L=Q, K=Q, T=3Q.
When Q=1000,
then L = 1000, K = 1000 and T = 3000
When Q = 8000
then, L = 8000, K = 8000, T = 24000
When Q = 20000
then L = 20000, K = 20000, T = 60000
1.c. If it took one hour of labour, half an hour of sewing time machine and two yards of cloth to make a shirt, the fixed proportion production function would be as follows :
Q = min(L, 2K, T/2)
2.a. When ONE shirt has to be produced i.e. Q = 1, the most efficient set of inputs is as follows :
L = 1, K = 1, T = 3.
Cost C(Q) = L(6) + K(1) + T(2) = 6+1+6=13.
Thus, cost of producing one shirt is $13.
When Two shirts have to be produced, i.e. Q = 2, the most efficient set of inputs is as follow:
L = 2, K = 2, T = 6
Cost C(Q) = 2(6) + 2(1) + (6)(2) = 12 + 2 + 12 = 26
Cost of producing two shirts is $26.
Thus, C(2) = 2(C(Q))
When Q = 1000, the most efficient set of inputs is :
L = 1000, K = 1000, T = 3000
C(Q) = 6(1000) + 1(1000) + 3(2)(1000) = 6000 + 1000 + 6000 = 13000
Thus, C(1000) = $13000
For Q shirts, L = Q, K = Q, T = 3Q.
C(Q) = 6L + 1K + 2T = 6Q + Q + 6Q = 13Q.
Thus, cost of producing Q shirts is $13Q.
2.b. Long run total cost = $13Q
Long run average total cost = Long run total cost/Q = 13Q/Q = $13
Thus, ATC = $13
2.c. MC = ?TC/?Q
TC(1) = 13 and TC(2) = 26
?TC = 26 - 13 = 13.
?Q = 2 - 1 = 1.
Thus, MC = 13/1 = $13.
Alternatively,
Since the market is in constant returns to scale, ATC = MC.
ATC = $13.
Thus, MC = $13
In a perfect competiton, MR = AR = PQ/Q = P = 30 - Q/1000
Thus, MR = 30 - Q/1000.
Given that for profit maximization, MR = MC = $13
30 - Q/1000 = 13
Q/1000 = 17
Q* = 17000
P* = 30 - Q*/1000
P* = 30 - 17000/1000
P* = 30 - 17
P* = $13
For optmal input levels,
L* = Q*, K* = Q*, T* = 3Q*
L* = 17000
K* = 17000
T* = 51000
Total Cost at profit maximizing level = 6(17000) + 1(17000) + 2(51000) = 102000 + 17000 + 102000 = $221000
Total Revenue at profit maximising level = 13(17000) = $221000
Profit = Total Revenue - Total Cost = $0
?* = $0
TFC = $25000
MC = 13 - 1 = $12
TVC = MC(Q) = 12(5000) = $60000
TC = TFC + TVC = $85000
For profit maximization MR = MC
MR = $12 = P
12 = 30 - Q/1000
Q/1000 = 18
Q* = 18000
P* = 30 - 18000/1000 = 30-18=12
But since K is restricted to 5000,Q* can only be 5000.
L* = K* = 5000
T* = 15000
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