A consumer has preferences for two goods that are represented by the utility function u(1, 2) = 2√₁ + √₂ if x₁ ≤ T. good 1; that is, if ₁ > For 2₁ > I, he gets no additional benefit from consuming more of then he is indifferent between (2₁, 2) and (T, x₂). (a) Show that this consumer's preferences are monotone and convex. Solution: If x'>x, then 2√√x₁+√√√₂ > 2√√₁+√√₂ since √ is strictly increasing. This implies that preferences are monotone for x1 <. For x1 > T, once can replace x1 with T. For convexity, note that, for x₁ ≤, indifference curves consist of solutions to 2√1 + √₂ = k for some k. Along the indifference curve, for x₁ < I we have > >0 and for x₁ > we have = 0. Since preferences are continuous and monotone, it follows that they are (weakly) convex. dz (b) Find the consumer's Hicksian demands for each good. Solution: Solving the expenditure minimization problem gives 21 = 2up2 P1+ 4p2/ and x2 = upi P1+ 4p2. 2 = provided that 21 ≤, i.e. provided that u ≤ (2+p₁/2p₂) √ (note that part (a) implies that the solution to the FOCs is indeed a minimum). Otherwise, the solution is x₁ = T and x2 = (u - 2√7)². Putting these together, we have 2up2 P1+4p₂ h(p. u) = {((²p)², (mm)²) ifu≤ (2+p₁/2p2)√VF, | (I, (u - 2 √F)²) otherwise.
A consumer has preferences for two goods that are represented by the utility function u(1, 2) = 2√₁ + √₂ if x₁ ≤ T. good 1; that is, if ₁ > For 2₁ > I, he gets no additional benefit from consuming more of then he is indifferent between (2₁, 2) and (T, x₂). (a) Show that this consumer's preferences are monotone and convex. Solution: If x'>x, then 2√√x₁+√√√₂ > 2√√₁+√√₂ since √ is strictly increasing. This implies that preferences are monotone for x1 <. For x1 > T, once can replace x1 with T. For convexity, note that, for x₁ ≤, indifference curves consist of solutions to 2√1 + √₂ = k for some k. Along the indifference curve, for x₁ < I we have > >0 and for x₁ > we have = 0. Since preferences are continuous and monotone, it follows that they are (weakly) convex. dz (b) Find the consumer's Hicksian demands for each good. Solution: Solving the expenditure minimization problem gives 21 = 2up2 P1+ 4p2/ and x2 = upi P1+ 4p2. 2 = provided that 21 ≤, i.e. provided that u ≤ (2+p₁/2p₂) √ (note that part (a) implies that the solution to the FOCs is indeed a minimum). Otherwise, the solution is x₁ = T and x2 = (u - 2√7)². Putting these together, we have 2up2 P1+4p₂ h(p. u) = {((²p)², (mm)²) ifu≤ (2+p₁/2p2)√VF, | (I, (u - 2 √F)²) otherwise.
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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