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Since at most two constraints can bind constraint

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Since at most two constraints can bind, constraint qualification is satisfied. The Lagrangian is L = x + y 2 - λ ( px + y - 6) + μx + νy . The first-order conditions are 0 = 1 - λp + μ 0 = 2 y - λ + ν. The first equation tells us that λ 1 /p > 0. Complementary slackness then implies px + y = 6. Now there are two cases to consider. If x = 0, then y = 6. By complementary slackness, ν = 0. The second first-order condition implies λ = 12. Provided 12 p 1, μ = 12 p - 1 0, and we have a critical point (0 , 6). If y = 0, then x = 6 /p . By complementary slackness, μ = 0. The first first-order condition implies λ = 1 /p . Substituting in the second equation, we find ν = 1 /p . This is our second critical point, (6 /p, 0). We now turn to the second-order conditions. Since there are n = 2 variables and m = 2 constraints, we have no second-order conditions available. What we can do is compare the values of utility: u (0 , 6) = 36 while u (6 /p, 0) = 6 /p . If p > 1 / 6, (0 , 6) is the maximizer while if p < 1 / 6, (6 /p, 0) is the maximizer. If p = 1 / 6, both are maximizers. 4. A consumer is endowed with T > 0 units of a resource r that may either be consumed or sold at price w > 0. The resource cannot be bought on the market, only sold. The consumer also consumes a consumption good c which has price p > 0. The budget constraint is pc + wr wT . The consumer is also subject to the constraints c 0, r 0, and r T (the last constraint reflects the fact that the resource cannot be purchased). The utility function is u ( c, r ) = ln c +2 ln r . Solve the consumer’s problem, paying attention to constraint qualification and the second-order conditions.
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MATHEMATICAL ECONOMICS FINAL, DECEMBER 10, 2002 Page 3 Answer: The Lagrangian is L = ln c + 2 ln r - λ ( pc + wr - wT ) + μc + νr - ρ ( r - T ). The first-order conditions are 0 = 1 c - λp + μ 0 = 2 r -
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