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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 Zekes 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 Yolondas marginal utility is positive. By taking the derivatives of the utilities, it can be shown that Xaviers 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 Xaviers utilty is increasing in h but Zekes is decreasing so we cant 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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