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Unformatted text preview: Solutions for Problem Set 2 Problem 1 Part a First we assume h ≥  . To check if an h is Pareto efficient we want to find the range of h where the derivates of the utility functions do not all have the same sign. Since Zeke’s utility function is v z ( h ) = h 2 and the derivative is v z ( h ) = 2 h < 0 for any h , we just need to find the range where Xavier or Yolonda’s marginal utility is positive. By taking the derivatives of the utilities, it can be shown that Xavier’s utility is increasing on h ∈ [0 , 3 2 ] and Yolonda increases on h ∈ [0 , 1 3 ]. Therefore for any h > 3 2 , v i (3 / 2) > v i ( h ) for all i , and thus no h > 3 2 can be Pareto efficient since all three could be made better off by setting h = 3 2 . For all h ∈ , 3 2 Xavier’s utilty is increasing in h but Zeke’s is decreasing so we can’t make one of the two better off without making the other worse off. So each h ∈ , 3 2 is Pareto Efficient. Part b The sum of the utilities is: v x ( h ) + v y ( h ) + v z ( h ) = 3 h h 2 + 2 h 3 h 2 h 2 = 5 h 5 h 2 Part c V ( h ) = 5 10 h = 0 → h * = 1 2 Part d First note that at h = 0, v x (0) = v y (0) = v z (0) = 0 and that ∑ i v i (0) = 0. Also, v x ( 1 2 ) = 5 4 , v y ( 1 2 ) = 1 4 , v z ( 1 2 ) = 1 4 . Using this requirement (ii) becomes:....
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This note was uploaded on 01/14/2012 for the course ECON 201 taught by Professor Witte during the Spring '08 term at Northwestern.
 Spring '08
 Witte
 Macroeconomics, Utility

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