400ps4-11-soln - ECON 400 Winter 2011 Solutions for Problem...

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ECON 400 Hartman Winter 2011 Solutions for Problem Set IV 1.a. Solve 1 ( ) a b pb K L w to get 1 1 1 1 1 1 1 ( , , ) ( ) a S b b b b L p w K b p w K . 1.b. Substitute * K for K in ( , , s L p w K ) to get 1 1 1 1 1 1 * 1 1 1 1 1 1 1 1 1 1 1 1 ( 1) 1 1 1 (1 )(1 ) 1 (1 )(1 ) 1 (1 )(1 ) (1 )(1 ) 1 1 1 1 1 [ , , ( , , )] [ ] b b b b a S b b b a b a b a b a b a b b ab a ab b a a a a a b a b b b a b b b a b b a b b a b a b a b a b a b L p w K p w r b p w a b p r w a b p w r a b p r w                             1 * 1 a a b L   2.a. Definition (i) is symmetric because 12 21 2 1 1 2 y y f f x x x x . Definition 2 is symmetric because * * 1 1 2 2 1 2 2 1 ( , , ) ( , , ) x w w p x w w p w w because of reciprocity. 2.b. The definitions are equivalent because * 1 12 2 2 11 12 12 ( ) x f w p f f f , * 2 21 2 1 11 12 12 ( ) x f w p f f f , and 12 21 f f . 2.c. No, because more than just the cross partial 12 21 f f will be involved. 3.a. The first order conditions are 1 1 1 2 ( ) ( ) 0 R x C x x t and 2 2 1 2 ( ) ( ) 0 R x C x x ; the second order (sufficient) conditions are 1 0 R C   , 2 0 R C   , and 2 1 2 ( )( ) ( ) 0 R C R C C      . [For the last condition, note that 2 2 ( ) ( ) C C   .] Solve the first order conditions for * 1 ( ) x t and * 2 ( ) x t , substitute these
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  • Fall '08
  • Ellis,G
  • Elementary algebra, dt, Rate equation, R1, R2

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