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 () ab pbK L w to get 11 1 111 1 (,, ) ( ) a S bbb b LpwK b p w K  . 1.b. Substitute * K for K in (,, s LpwK ) to get 1 1 * 1 1 1 1 1 1 1 ( 1 ) 1 (1 )(1 ) 1 (1 )(1 ) 1 (1 )(1 ) (1 )(1 ) 1 [,, (,,) ] [ ] bb b b a S a ba b a b b ab a ab b a a aa a b b b b b b a b b LpwK pw r b p w a b p r w p w r prw        1 * 1 a L 2.a. Definition (i) is symmetric because 12 21 21 12 y y ff x xx x      . Definition 2 is symmetric because ** 112 212 (, ,) xww p ww because of reciprocity. 2.b. The definitions are equivalent because * 2 2 1 1 2 1 2 xf wp f  , * 22 1 2 1 1 2 1 2 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 2 ( ) 0 R xC x xt  and ( )0 Rx Cx x  ; the second order (sufficient) conditions are 1 0 RC   , 2 0 , and 2 ( ) 0 C . [For the last condition, note that ( ) CC  .] Solve the first order conditions for * 1 x t and * 2 x t , substitute these back into the first order conditions to get identities, and differentiate through the identities with respect to t to get the following system: 1 2 1 0 dx dx C dt dt dx dx CR C dt dt To solve for * 1 / dx dt , multiply through the first equation by 2 R C and the second equation by C and then add the results together to get * 2 1 2 [( )( ) ( ) ] dx R CR C C R C dt  ; this can be solved to give * 2 0.
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400ps4-11-soln - ECON 400 Winter 2011 Solutions for Problem...

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