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Wednesday, 28 November 2012

Price Theory

Question A Part 1. 1 1 maxf x1 + x2 setoff Order Conditions (FOC): x1 x2 x1 x2 1 1 1 1 [p1 x1 + p2 x2 Ig = = p1 p2 = = p1 p1 = p2 p2 p1 p2 1) Let r x1 x2 x1 x1 = ( r)1=( = x2 ( r)1=( 1) I p1 x1 ]( r)1=( = [ p2 I = [ rx1 ]( r)1=( p2 = ( r)1=( 1 + ( r)1=( 1=( 1) 1) 1) 1 = 1 x1 x2 1=( 1) = I ( r)1=( 1) r I r = = 1=( 1) 1) r p 1=( 1) r p 1 + ( r) 2 1 1+ r I p1 = x1 ( r) = 1) 1 1 + ( r)1=( 1) r I p2 Cobb-Douglas check. Set x1 = = 0: I 1 = p2 p1 1 + I 1 ( r) 1 = 1 + ( r) 1 r p2 r1+ so that the expenditure on trade good 1 is a …xed divide ( =(1 + )) of the value of the endownment (which must be the case for Cobb-Douglas). Do we necessitate to worry about corner solutions? No, unless = 1 (perfect substitutes). This is because the MRS (how much much of 2 the single(a) requires for giving up a whole of 1) is x1 x2 1 1 which, for < 1, is in…nite for x1 = 0, (meaning that for any positive price of good 1, the individual will want *some* of good 1), and postal code for x2 = 0 (meaning that for any positive price of good 2, the individual will want *some* of good 2). 1 Part 2.
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minfp1 x1 + p2 x2 FOC: p1 p2 = = x1 x2 1 1 1 1 f x1 + x2 U0 g x1 x2 x1 1 1 1 = = p1 p2 rx2 r 1 1) 1) x1 x1 But x1 + x2 ( r) =( 1) = ( r)1=( = ( r) =( x2 x2 = = = = [ U0 U0 + ( r) + ( r) =( 1) x2 + x2 x2 x2 x1 U0 ]1= U0 1= =( 1) = ( r)1=( = [ 1) x2 = ( r)1=( 1) 1) 1) [ + ( r) =( 1) ]1= U0 1= ( r) =( + ( r) =( ]1= U0 1= Part 3. 1 1 maxf x1 + x2 FOC: x1 x2 1 1 [p1 x1 + p2 x2 [p1 e1 + p2 e2 ]]g = = p1 p2 2 x1 x2 1 1 = = = = = x1 x2 x1 x1 Let r x1 x2 = = = = p1 p1 = p2 p2 p1 1=( 1) ( ) p2 p1 1=( 1) x2 ( ) p2 p1 e1 + p2 e2 p1 x1 p1 1=( 1) [ ]( ) p2 p2 p1 p1 p1 1=( 1) [ e1 + e2 x1 ]( ) p2 p2 p2 p1 p2 ( r)1=( 1) [re1 + e2 ] 1 + ( r)1=( 1) r p1 1=( 1) x1 ( ) = x1 ( r) 1=( 1) p2 1 [re1 + e2 ] 1 + ( r)1=( 1) r = 0:... If you want to get a full essay, order it on our website: Ordercustompaper.com

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