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Lecture

# Classroom Lecture Notes - Kings

5 Pages
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School
Department
Economics
Course
Economics 2150A/B
Professor
Hugh Cassidy
Semester
Fall

Description
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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