answers3

Mathematics for Economists

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ANSWERS PAMPHLET 101 and ] a j f ( x p ( b , a )) 5 L a j ( x p , b p , a p , l p , m p ) . 19.10 Form L ( x ,a, ) 5 f ( x ) 2 j h j ( x ,a ). Let ( x p ( a p ) , p ( a p ) ) denote the maximizer and its multiplier when the parameter a 5 a p . a 7→ h j ( x p ( a ) ) is the zero function. f ( x p ( a )) 5 f ( x p ( a )) 2 X p j ( a ) h j ( x p ( a ); a ) 5 L ( x p ( a ) , p ( a ); a ) for all a. d da f ( x p ( a )) 5 d da L ( x p , p ( a ); a ) 5 X i L x i ( x p ( a ) , p ( a ) ) dx p i da ( a ) 1 X j L ]m j ( x p ( a ) , p ( a ) ) d p j da ( a ) 1 L a ( x p ( a ) , p ( a ) ) 5 0 1 X j ( 2 h j ( x p ( a ) )) d p j da ( a ) 1 L a ( x p ( a ) , p ( a ) ) 5 L a ( x p , p ( a ) ) . 19.11 max f ( x ) subject to h 1 ( x ) 5 a 1 ,...,h k ( x ) 5 a k . Suppose the constraint qualification holds and that the solution x p depends on a 5 ( a 1 ,...,a k ). The Lagrangian is L ( x , ; a ) 5 f ( x ) 2 i ( h i ( x ) 2 a i ). By Theorem 19 . 5, a j f ( x p ( a )) 5 L a j ( x p ( a ) , p ( a ); a ) . It follows from the above formula for L that L a j 5 j .So , a j f ( x p ( a )) 5 j ( a ). But this is the conclusion of Theorem 19 . 1 . 19.12 max x 2 1 y 2 subject to x 2 1 xy 1 . 9 y 2 5 3 .
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102 MATHEMATICS FOR ECONOMISTS Write the constraint as x 2 1 xy 1 by 2 5 3. L 5 x 2 1 y 2 2 l ( x 2 1 1 by 2 2 3) . ] L b 52 y 2 . For b 5 1, the max is at ( x, y, ) 5 ( 6 p 3 , 7 p 3 , 2) with f p 5 6; and the min is at ( x, y, ) 5 ( 6 1 , 6 1 , 2 6 3) with f p 5 2. For b 5 0 . 9, f p < 6 1 ( 2 2 ? 3) ? ( 2 . 1) 5 6 . 6 at the max and f p < 2 1 ( 2 (2 6 3) ? 1) ? ( 2 . 1) 5 2 1 (2 6 30) 5 2 1 15 at the min. If we want actual distance and not distance squared, we would take the square roots of these numbers and obtain 2.569 and 1.438, respectively. 19.13 max x 2 1 x 1 ay 2 subject to 2 x 1 2 y # 1 ,x $ 0 ,y $ 0 . The Lagrangian is L 5 x 2 1 x 1 2 2 1 (2 x 1 2 y 2 1) 1 2 x 1 3 y. For a 5 4, x p 5 0, y p 5 0 . 5, p 1 5 2, p 2 5 3, p 3 5 0, and f p 5 1. At these values, L a 5 y 2 5 0 . 5 2 5 0 . 25 so f p (4 . 1) < f p (4) 1 L a ? D a 5 1 1 0 . 25(0 . 1) 5 1 . 025 . 19.14 18.2): H 5 02 x 1 yx 1 2 y 2 x 1 y 2 2 2 2 x 1 2 y 2 2 2 2 5 033 32 6 3 2 2 6 3 3 2 2 6 6 3 at the minimizer (1 , 1 , 2 6 3). det H 24 , 0, the SOC for a constrained min. At the max, ( p 3 , 2 p 3 , 2), det H 5 det 0 p 3 2 p 3 p 3 2 2 2 2 2 p 3 2 2 2 2 51 24 . 0 , the SOC for a constrained max.
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ANSWERS PAMPHLET 103 18 . 3): det H 5 det 02 x 2 1 2 x 2 1 2 l 0 2 10 2 52 (2 1 2 1 8 x 2 ) , 0 , the SOC for a constrained min. 18 . 5): 00311 00111 31200 11020 11002 has positive determinant, the SOC for a constrained min when there are 3 variables and 2 constraints. 19.15 If ] h x ( x p ,y p ) ± 0, then C h can be written as x 5 c ( y ) around ( x p p ); i.e., h ( ( y ) ) ; 0 for all y near y p and 0 ( y ) h y ( ( y ) ) ` h x ( ( y ) ) . Let F ( y ) 5 f ( ( y ) ) .So , F 0 ( y p ) 5 0and F 00 ( y p ) , 0 implies that y p is a strict local max of F and that ( x p p ) is a strict local constrained max of f . F 0 ( y p ) 5 f x ( ( y p ) p ) 0 ( y p ) 1 f y ( ( y p ) p ) 5 f x ( ( y p ) p ) 0 ( y p ) 1 f y ( ( y p ) p ) 2 m p h x ( ( y p ) p ) 0 ( y p ) 1 h y ( ( y p ) p ) 5 L x ( x p p ) 0 ( y p ) 1 L y ( x p p ) 5 0 , since L x 5 L y 5 0byFOCs . As in the proof of Theorem 19 . 7, F 00 ( y p ) 5 L xx ( 0 ) 2 1 2 L xy 0 1 L yy 1 L x 00 5 L xx ± 2 h y h x 2 1 2 L ± 2 h y h x 1 L yy 1 0 5 1 h 2 x h 2 y L xx 2 2 h x h y L 1 L yy h 2 x · , which is negative by hypothesis b of the Theorem.
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104 MATHEMATICS FOR ECONOMISTS 19.16 The proof is basically the same, except that hypothesis b now implies that F 00 ( x p ) . 0. The conditions F 0 ( x p ) 5 0and F 00 ( x p ) . 0 imply that x p is a
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answers3 - ANSWERS PAMPHLET 101 and aj f (x (b, a) L (x , b...

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