Example Cobb-Douglas production function

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					Example: Cobb-Douglas production function. f(L,K) = Y = 3 K⅓L⅓. Cost: Labor is $9 and Captal is $27. Y = 3K 3 L 3 ⇔ 1 Y Isoquants: K 3 = 1 ⇔ 3L 3 Y3 K= 27 L To produce 3 units, we have K = 1/L. To produce 1 unit we have K=1/27L, to produce 6 units we have Y = 8/L.
Isoquants
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 L 6 7 8 9 10
1 1

Y=6 Y=3 Y=1

C 1 − L . So for $200, we get 3 27 K = 7.4 - ⅓L, and for $100 we get K = 3.7 - ⅓L. Drawing, we get:

Isocost curves: C(K,L) = 9L + 27K, so K =

K

Isoquants and Isocost curves
9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 L 6 7 8 9 10

Y=6 Y=3 Y=1 C = 200 C = 100

K

Recall that to minimize cost for a given quantity, we had to set MRTS=wL/wK. Suppose we’re producing 6 units. What is the minimum cost? ∂f ( L, K ) 1 −2 3 3 K ∂L = K L so we need MRTS = = 1 −2 ∂f ( L, K ) L K 3L 3 ∂K K 9 1 = ⇔K= L 3 L 27 What now? We wanted to produce 6 units, so 6 = 3 K⅓L⅓ = 3 (⅓L) ⅓L⅓ = 3 (⅓) ⅓ L⅔. Solving for L, we get L =
2 3

2 3
1 3

⇔L=

2

3 2 1

= 4.9 and K = ⅓L = 1.6 and the total cost is

12 3

12 = 88.18 3 11 2 1 3 3 In general, suppose we want to find the cost as a function of the quantity produced. K and the price ratio is ⅓, so K=⅓L and Q = 3 K⅓L⅓ = 3 (⅓L) ⅓L⅓ = 3 (⅓) ⅓ MRTS = L C =9 2
1 2

3 2

+ 27

3 2

⎛ ⎜ ⎜ Q L⅔. So L = ⎜ 1 ⎜ 1 3 ⎜ 3⎛ ⎞ ⎜ ⎜ 3⎟ ⎝ ⎝ ⎠

⎞ ⎟ ⎛ ⎟ ⎜Q ⎟ =⎜ 2 ⎜ 3 ⎟ ⎝3 ⎟ ⎟ ⎠
3 2 3

3 2

⎞2 Q2 Q2 ⎟ and K = So = ⎟ 9 3 ⎟ ⎠
3 3

3

3

3

Q Q + 27 = 3Q 2 + 3Q 2 = 6Q 2 3 9 Long-run cost function. C (Q) = 9
Now, what happens if K is fixed at 1.6 and now we wish to produce 3 units? (SHORT RUN) Then we determine how much L we need to produce 3 when K is fixed at the previous level: 3 = 3
⎛ ⎜ 3 1 1 ⎜ 22 3 = 3K 3 L3 = 3⎜ 1 ⎜ 12 ⎜ 3⋅ ⎝ 3 ⎞3 ⎟ 1 ⎟ 1 1 3 3 ⎟ L ⇔L = ⎟ ⎛ ⎟ ⎜ 3 ⎠ ⎜ 22 ⎜ 1 ⎜ 12 ⎜ 3⋅ ⎝ 3
1 1 ⎛ ⎜ 3⋅ 1 2 ⎜ =⎜ 3 3 1 ⎞3 ⎜ 22 ⎜ ⎟ ⎝ ⎟ ⎟ ⎟ ⎟ ⎠

3 2

⎞3 1 ⎟ ⎟ 32 ⎟ ⇔L= 3 ⎟ 22 ⎟ ⎠

1

Then C = 9

3 2

1 2 3 2

+ 27

2

3 2 1

= 48.71

12 3⋅ 3
3 2

In the long run we could have produced this at a cost of C (3) = 6 ⋅ 3 = 31.18 In general we can find the short run cost function like this: We fix K at ĸ. Then to 1 1 1 Q3 Q Q3 3 3 3 produce Q units we have: Q = 3 k L ⇔ L = 1 ⇔ L = Then C = 9 + 27 k . 27 k 27 k 3 3k