Cobb-Douglas: Given Q*, ie. Isoquant, the goal is to find the lowest isocost line that reaches the Q*
isoquant.
Example
Total cost line can reach Q* but it’s not the cheapest
Total cost line 2 can't reach Q* so we can't afford it
Total cost line 3 is the lowest isoquant line, cheapest way to produce Q*
Solving for L and K
2 steps:
1) Tangency condition – Slope of the isoquant = slope iso cost line
- MRTS =L,K /MP =Lw/r K
Tangency condition
- MP /L = MP /r K Equate marginal products per dollar
2) Substitute into isoquant Q* = F(L,K)
Sub in solution from step 1 into Q*, then solve for L or K
Example
1/2 1/2 1/2 1/2 1/2 1/2
Q = 2L K MP =LK /L MP =KL /K
MRTS = MPL/MPK = K /L /L /K2 1/2 1/2 1/2= K/L
K/L = w/r K = (w/r)L
1/2 1/2
Sub K = (w/r)L into Q* = 2L K
1/2 ½ 1/2 1/2 1/2 1/2
Q* = 2L (w/r L) = 2L (w/r) L = 2(w/r) L
1/2
L = Q*/2(w/r)
Q* = 4, r = 4, w = 1
L = 4/2(1/4) 1/2
L = 4
K = (w/r)L
K = (1/4)4
K = 1 (L*,K*) = (4,1) Cost of the bundle: wL + rK
1(4) + 4(1) = 8
Perfect Substitutes (Linear) Q = al + bK, MPL= a, MPK= b
- Check marginal product per dollar condition
1) MP /wL> MP /r K Therefore optimal bundle will include K* = 0
o Q* = aL* + bK*
o Q* = aL* + b(0)
o L* = Q*/a
2) MP /w < MP /r Therefore optimal bundle will include L* = 0
L K
o Q* = aL* + bK*
o Q* = a(0) + bK*
o K* = Q*/b
3) MP /wL= MP /r K Any L* and K* combination such that Q* = aL*+bK*
Example
Q* = 10, w= 3, r = 1, Q = 2L+K
MP = 2/3 < MP = 1/1 = 1
L K
L* = 0 therefore
Q* = 10 = 2(0)* +K*
10 = K*
Example 2
Q* = 10, w= 2, r = 1, Q = 2L+K
MP L 2/2 = 1 = MP K 1/1 = 1
Therefore any (L*,K*) such that, 10 = 2L* + K* is optimal
Calculate cost by choosing any (L*,K*) bundle that produces Q* = 10
Set K*=0, L* = 10/2 = 5. TC = wL* +rK*
TC = 2(5) + 1(0)
TC = 10 Perfect Compliments (Fixed Proportion): Q = min {aL,bK}
- Optimality Condition aL = bK
- Substitute aL* = bK* into production function
- aL = bK L = (b/a) K, Substitute

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