Using the data table:
Assuming that the price of labor is w = 10 and the price of capital is k=40, use Table 1 to do the following:
If capital and labor are complements, a decrease in K always
leads to a decrease in L, since capital increases the incremental
returns to labor. In contrast, when capital and labor are
substitutes, a decrease in K leads to a weak increase in L, and so
capital and labor will move in opposite directions. From the table,
we can see that they are substitutes.
The Cobb-Douglass production function is of the form: Y =
KALB Where Y=output, K=amount of capital,
L=amount of labor. Again, from the table, we see that increase in K
and L by factors of 2 change the output by values to some power of
2. This indicates a Cobb-Douglass production function.
If A+B = 1, the production function exhibits constant returns to
scale. If A+B >1, the production function exhibits increasing
returns to scale If A+B <1, the production function exhibits
decreasing returns to scale.
If we use Y=KA LB to understand whether the
production exhibits increasing, decreasing or constant returns to
scale, we want to look at how 2Y compares to Y(2K,2L) =
(2K)A (2L)B . Does doubling both inputs
exactly double output? If (2K)A (2L) B >
2Y the production function exhibits increasing returns to scale. If
(2K)A (2L)B < 2Y the production function
exhibits decreasing returns to scale. If (2K)A
(2L)B = 2Y the production function exhibits constant
returns to scale. This data gives constant returns to scale since
for K =1, L = 4, Y = 1.00. For K=2,L=8, Y=2.00.
APL = Q/L and MPL is the change in APL due to increase in L by
1.
K | L | Q | APL | MPL |
4 | 0 | 1.00 | - | - |
4 | 1 | 1.56 | 1.56 | 1.56 |
4 | 2 | 1.83 | 0.915 | 0.27 |
4 | 3 | 2.05 | 0.683 | 0.22 |
4 | 4 | 2.25 | 0.5625 | 0.20 |
4 | 5 | 2.43 | 0.456 | 0.18 |
4 | 6 | 2.60 | 0.433 | 0.17 |
4 | 7 | 2.76 | 0.394 | 0.16 |
Next, we compute Table 3. Fixed cost is cost of capital (4*40 =
160). Variable costs are costs of labor.
MC = change in TC/change in Q
AFC = FC/Q
K | L | Q | FC | VC | TC | MC | AFC | AVC | ATC |
4 | 0 | 1.00 | 160 | - | 160 | - | 160 | - | |
4 | 1 | 1.56 | 160 | 10 | 170 | 17.86 | 102.56 | 6.42 | 108.98 |
4 | 2 | 1.83 | 160 | 20 | 180 | 37.04 | 87.43 | 10.93 | 98.36 |
4 | 3 | 2.05 | 160 | 30 | 190 | 45.45 | 78.05 | 14.63 | 92.68 |
4 | 4 | 2.25 | 160 | 40 | 200 | 50 | 71.11 | 17.78 | 88.89 |
4 | 5 | 2.43 | 160 | 50 | 210 | 55.55 | 65.84 | 20.58 | 86.42 |
4 | 6 | 2.60 | 160 | 60 | 220 | 58.82 | 61.54 | 23.08 | 84.62 |
4 | 7 | 2.76 | 160 | 70 | 230 | 62.5 | 57.97 | 25.36 | 83.33 |
4 | 16 | 4.00 | 160 | 160 | 320 | 72.58 | 40 | 40 | 80 |
4 | 32 | 5.83 | 160 | 320 | 480 | 87.43 | 27.44 | 54.88 | 82.32 |
4 | 64 | 9.00 | 160 | 640 | 800 | 100.95 | 17.78 | 71.11 | 88.89 |
Now for Table 4, for each value of Q we look at cost minimizing
values in Table 1.
Thus, for Q = 1, we have K = 1 and L = 4 as minimum cost option,
giving TC = 80.
For Q = 2, K = 2, L = 8. Thus, TC = 160.
For Q = 4, K = 4, L = 16, thus, TC = 320.
If we sell each unit at 73,
Q = 1 ] 73 - 80 = -7
Q = 2 ] 146 - 160 = -14
Q = 4 ] 292 - 320 = -28
Hope this helped!
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