1. Production and Cost: One Variable Input
Chapter 6
Production and Cost:
One Variable Input

2. The Production Function
We begin by assuming that a firm produces one good, Y, using 2 inputs called input 1 and input 2.
Let y = quantity of good Y
z1 = quantity of input 1
z2 = quantity of input 2

3. The input bundle is (z1, z2).
Example: if the firm uses 10 units of input 1 and 6 units of input 2, the input bundle is (10, 6).

4. How inputs are combined to produce quantities of output varies depending on the technology the firm uses for production.
We assume that there is one function that is technically efficient = it maximizes the qty of output that can be produced from a given bundle of inputs.

5. We use the technically efficient technology to define:
The production function
y = F(z1, z2)
It tells us the maximum qty of good Y that can be produced from any input bundle (z1, z2).

6. Fixed-Proportions Production Function
= the inputs are always used in a fixed ratio
Example: one nut + one bolt for one fastener
Example: 1 oz vodka + 6 oz orange juice + 2 oz ice for one screwdriver
Note that the fixed proportion doesn’t have to be 1:1.

7. Example
To make the best cranapple drink, need to use 150 ml of apple juice and 100 ml of cranberry juice.
Let y = number of cranapple drinks
z1 = ml of apple juice
z2 = ml of cranberry juice

8. Say you had z1 = 600 ml of apple juice and z2 = 500 ml of cranberry juice.
You have enough apple juice to make
600 ml = 4 drinks i.e., z1 / 150
150 ml
You have enough cranberry juice to make
500 ml = 5 drinks i.e., z2 / 100
100 ml

9. The most cranapple drinks you could make is 4.
You don’t have enough apple juice to make any more.
The fixed-proportions production fn is
y = min ( z1/ 150, z2/ 100 )

10. Functions of this type are known as Leontief production functions.
Let’s do another example using more inputs.

11. Variable-Proportions Production Function
= the input mix can vary
You can substitute by increasing the amount used of one input and decreasing the amount used of another input.

12. Example: A Courier Service
The courier service has one delivery truck.
Let y = courier services measured in km
z1 = driver’s time measured in hours
z2 = gasoline measured in litres

13. How many km driven depends on varying combinations of how many hours the driver drives and how much gas he uses.
At a higher speed, the driver drives fewer hours but more gas is used up to cover a given distance.

14. So, maximum y for a given z1 hours of driving time is determined by how fast the truck is driven.
Let s = speed in km per hour
Then, y < sz1 (speed per hr times # of hrs)

15. But, we need to know how much gas we use to drive at speed s.
Suppose we know: the relationship is
Km / litre of gas = 1200 s

16. This means that y < z2 s
For example, if the speed is 120 kph, on 20 litres of gas you can’t go more than 200 km.

17. To find the courier’s production fn, we need to choose the speed s
To find the courier’s production fn, we need to choose the speed s* that maximizes distance y*.
Let’s do it diagrammatically.
Plot y < sz1 and y < z2 to find s* and
s y*.
Assume that (z1, z2) is a given input bundle (pretend they are constants).

19. The shaded area satisfies both equations.
We max y at y* at a speed of s*.
The point at which the 2 constraints intersect determines y*.
Let’s solve for it:

20. 1200 z2 = sz1 s
1200 z2 = s2z1
s* = ( 1200z2 / z1 )1/2
Substitute this equation for s* into either constraint (easier to use y* = sz1 )

21. y* = ( 1200z2 / z1 )1/2 z1
y* = ( 1200z2)1/2 z11/2
y* = (1200z1z2)1/2
The production fn is F(z1, z2)=(1200z1z2)1/2

22. This function is an example of a Cobb-Douglas production function.
The general form is:

23. Production Costs We need to distinguish between:
Sunk Costs: unavoidable, non-recoverable costs
Example: a retailer installs built-in shelves that can’t be removed no matter what happens to the business down the road.
Avoidable Costs: can be recovered (at least in part)
Example: a retailer buys free-standing shelves that he can resell and recoup his costs.

24. We also distinguish between:
Fixed Costs: do not vary with output
Examples: loan payments, rent…
Variable Costs: change when output levels change
Examples: labour, raw material costs…

25. Cost Minimization
We assume that the firm’s objective is to maximize profit = total revenue – total costs.
This implies that profit is maximized when costs are minimized.

26. LR Cost Minimization
By long run, LR, we mean the time horizon long enough so that a firm can vary all its inputs.
Suppose a firm wants to produce y units of output and in the LR can choose the qty of both input 1, z1 and input 2, z2 in one time period.

27. Let w1 = price of input 1
w2 = price of input 2
The total cost of any input bundle (z1, z2) is:
TC = w1 z1 + w2 z2

28. The LR cost minimization problem, then, is
min w1z1 + w2z2
z1 , z2
st y = F(z1, z2)

29. That is, the firm chooses input qty (z1, z2) that minimizes total costs subject to the constraint that it can actually produce y units of output.
Keep in mind that the input costs must be measured for the same unit of time – weekly, monthly, annually – whatever.

30. SR Cost Minimization
In the short run, SR, at least one input is fixed in quantity.
In the case of 2 inputs where the qty of input 2 is fixed at some number, the problem is to choose z1 to minimize the costs of producing y units of output.

31. Production with One Variable Input
The total product function, TP gives us the total output for any qty of the variable input given the fixed qty of the other inputs.
If z2 is fixed at z2 , the total product fn is
TP (z1 ) = F(z1 , z2 )

32. The Typical TP Function
The slope is different at different values of z1 C

33. The rate at which y changes as the qty of variable input z1 changes given the fixed qty of all other inputs is the marginal product of z1 , MP( z1 ).
MP( z1 ) = slope of TP( z1 )
= d F(z1 , z2 )
d z1

34. Total product is maximized at point C on the diagram where MP( z1 ) = 0.
The free-disposal assumption suggests that if a firm has more than the output-maximizing level of z1 , it will decide to use exactly that amount and no more (it will dispose of the additional units of input).

35. This means that MP cannot be < 0 and the segment CE on our diagram is the relevant part of the TP function.

36. C E

37. Notice that in our TP function, the function gets steeper until point A, and then starts to flatten until it reaches maximum TP at point C.
Point A is the point where diminishing MP sets in.
This is equivalent to the point where MP is maximized: we’d set
d MP( z1 ) = 0
dz1

38. Maximum TP where MP = 0
C E
Point of diminishing MP

39. Example Let y = total number of fasteners produced
z1 = quantity of nuts
z2 = quantity of bolts
z2 is fixed at 10 units.
The production function is
TP( z1 ) = z1 if z1 < 10
10 if z1 > 10

40. 10 10 1 MP 10

41. Average Product AP( z1 ) = TP( z1 ) z1
This is the average amount of output per unit of the variable input.

42. Example If TP( z1 , z2 ) = (1200z1z2)1/2 and z2 = 12
AP ( z1 ) = (14400z1)1/2 z1
= 120z11/2
So, AP ( z1 ) = 120z1-1/2

43. Max AP and AP = MP

44. We can derive AP and MP from the TP function.

45. Note these 3 things:
When MP > AP, AP is rising
When MP < AP, AP is falling
When MP = AP, AP is at a maximum.

46. Costs of Production: One Variable Input
If the qty of input 2 is fixed at z2 and the price of variable input 1 is w1, the cost minimization problem is:
min w1z1 z1
st y = TP (z1)
Choose z1 to minimize the expenditure on input 1, w1z1

47. Once we know how much output, y, a firm wants to produce, we can find from the TP function the minimum qty of the variable input needed to produce that amount.
Then, we can find the minimum variable cost.
VC (y) = minimum variable cost of
producing y units of output

48. Example: Our Courier The production fn is TP(z1z2) = (1200z1z2)1/2
z1= hours driven
z2= gas in litres
Suppose gas is fixed at 48 litres.
TP(z1) = (1200z1(48))1/2
= (57600z1)1/2
So, y = 240z11/2

49. If we wanted the truck driven 480 km, the minimum amount of time needed is
480 = 240z11/2
2 = z11/2
z1 = 4 hours
If the driver’s wage is w1 per hour, then the
VC(480) = 4w1

50. More generally, solve the TP function for z1* :
y = 240z11/2
y2 = 57600z1
z1* = y2___
57600

51. So, the VC when the courier has 48 litres of gas is:
VC(y) = w1z1* = w1y2
57600

52. Note that the VC function is, in essence, the inverse of the TP function.
Figure 6.7 on page 210 in the textbook goes through the full derivation if you are interested.
Anyway, the VC function looks like this:

54. Average variable cost, AVC, is the variable cost per unit of output.
AVC (y) = VC (y) y

55. The SR marginal cost, SMC (y), is the rate at which cost increases in the SR as output increases.
It is the slope of VC (y).
SMC (y) = VC’(y) = d VC(y) dy
Diagrammatically,

56. AVC is at a min at point B (the flattest possible ray)
AVC is at a min at point B (the flattest possible ray). The slope of ray OB = slope of the VC curve = SMC. SMC = AVC at min AVC. B

57. From the previous diagram, we can derive our typical SR cost curves:
SMC
AVC

58. 3 points:
When SMC < AVC, AVC decreases as y increases.
When SMC > AVC, AVC increases as y increases.
When SMC = AVC, AVC is at a minimum.

59. Average Product and Average Cost
Suppose y’ = TP (z1’) so AP (z1’) = TP (z1’)
z1’
The cost of z1’ = w1 z1’
Then, AVC (y’) = w1 z1’ = w1 z1’ = w1 TP (z1’)
y TP (z1’) z1’
= w1 AP (z1’)

60. So, AVC of an input = price of the input
average product of the input
That is, AVC is the inverted image of the AP function.

61. Marginal Product and Marginal Cost
Suppose we increase z1 by a small amount, D z1
The increase in VC, Dc, will be
Dc = w1 Dz1
The increase in output, Dy, will be approximately the MP of the input times the additional qty of variable input
Dy = MP (z1) D z1

62. But, SMC is the rate of increase in VC as y increases, that is, Dc Dy
So, SMC = Dc Dy = w1 Dz1 MP (z1) D z1
SMC = w1 MP (z1)
Then, SMC = price of the input
MP of that input
That is, SMC is the inverted image of the MP function.

63. Other Costs Fixed Cost, FC Given fixed input z2 with a price of w2 ,
FC = w2 z2
Average Fixed Cost, AFC
AFC (y) = FC / y
SR Total Cost, STC
STC (y) = VC (y) + FC

64. SR Average Total Cost, SAC
SAC (y) = STC(y) / y
Also, SAC(y) = AVC(y) + AFC(y)
Note: feel free to use the notation we learned in first year: MC or SRMC for marginal cost and ATC or SR ATC for average total cost.

65. SMC = dTC/dy = dVC/dy THE WHOLE PICTURE
Note that the distance between STC and VC is just FC.
Thus, SMC = slope of VC
= slope of TC
SMC = dTC/dy = dVC/dy

66. Numerical Example A firm faces the production function y = z12/3z21/3
z2 is fixed at 1
The price of input 1 is $5
The TP function is
TP(z1) = z12/3

67. The AP function is
AP(z1) = z12/3 z1
= z1-1/3
The MP function is
MP(z1) = d TP(z1) d z1
= 2 z1-1/3 3

68. To find the VC function, TP is y = z12/3 so re- arranging, we get z1 = y3/2
VC(y) = w1z1 = w1y3/2
Since w1 = 5, VC(y) = 5y3/2
If the firm wants to produce 120 y,
VC(120) = 5(1203/2) = $