400ps7-11-soln - ECON 400 Winter 2011 Solutions for Problem...

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E C O N 4 0 0 H a r t m a n Winter 2011 Solutions for Problem Set VII 1. The Lagrangian for this problem can be written as ( ) ab rK wL Q K L  L . The first order conditions are 1 0 K r aK L  L/ , 1 0 L w bK L   , and 0 QKL . 2. 21 1 1 11 2 1 (1 ) ) 0 0 a b KK KL K LK LL L a b KL a a K L abK L aK L abK L b b K L bK L aK L bK L     LLL LL L L . To see that the inequality holds, expand the determinant down the last column to get 1 1 2 1 2 2 1 ) ) 0 ) ) [ a b a b a b a b a b a b a b a a K L abK L aK L abK L b b K L bK L aK L bK L abK L b b K L a a K L abK L aK L bK L aK L bK L aK L bK L aK L a        221 22 1 2 22 3 23 2 2 2 2 2 ) ] [ ) ] [( 1 ) ( 1 ) ] [ ] 0 a b b a b bKL b K L a b aKL a b K L ab abb ab a K L ab ab    where the inequality follows because 0 a b rw aK L bK L  from the first order conditions. 3. The first two of the first order conditions imply that 1 1 ra K L a L wb K L b K  or that br LK aw . Substitute this into the constraint to get b b b b a b br QK K ab rwK aw    , and this can be solved to give 1 ˆ bb b b ab ab ab ab ab KabrwQ  . Similarly, the first two first order conditions imply that aw br , which can be substituted into the constraint to get a ba a a a
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400ps7-11-soln - ECON 400 Winter 2011 Solutions for Problem...

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