This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem Set 10 Solutions Economics 6170 TA: Christopher Handy December 10, 2008 Problem 1 (a) Define the set X = R 4 ++ and define the functions f , g 1 , and g 2 , each from X to R , by f ( x ) = x 1 2 1 + x 1 2 2 for all x X g 1 ( x ) = px 3 + x 4 px 1 x 2 for all x X g 2 ( x ) = 1 x 2 3 x 2 4 for all x X A pair ( x, ) X R 2 + satisfies the KuhnTucker conditions for problem ( S ) if it satisfies the following: 1 2 x 1 2 1 + 1 ( p ) = 0 (1 . 1) p x 3 + x 4 p x 1 x 2 (1 . 5) 1 2 x 1 2 2 + 1 ( 1) = 0 (1 . 2) 1 x 2 3 x 2 4 (1 . 6) 1 ( p ) + 2 ( 2 x 3 ) = 0 (1 . 3) 1 ( p x 3 + x 4 p x 1 x 2 ) = 0 (1 . 7) 1 + 2 ( 2 x 4 ) = 0 (1 . 4) 2 ( 1 x 2 3 x 2 4 ) = 0 (1 . 8) Since the objective function is increasing in x 1 and x 2 , we expect that the first constraint in problem ( S ) will hold with equality at any solution. In addition, the greater is the right hand side of the first constraint, the greater is the value of the objective function that we can achieve. So we expect the second constraint in problem ( S ) to hold with equality, also. Seeking contradiction, suppose that 1 = 0. Then (1.1) and (1.2) cannot hold, so we have a contradiction. Therefore 1 &gt; 0. By (1.7), then, we have p x 3 + x 4 p x 1 x 2 = 0 (1.9) 1 Again seeking contradiction, suppose that 2 = 0. But then (1.4) implies that 1 = 0, which contradicts what we have just shown. Therefore 2 &gt; 0, so by (1.8) we have 1 x 2 3 x 2 4 = 0 (1.10) From (1.3) and (1.4), we have that p = x 3 x 4 . Using this in (1.10), we have that ( p x 4 ) 2 + x 2 4 = 1, which gives x 4 = 1 p 1 + p 2 , x 3 = p x 4 = p p 1 + p 2 Now, using (1.1) and (1.2) to eliminate 1 , we have p = q x 2 x 1 , or x 2 = p 2 x 1 . Then we can use (1.9) and p = x 3 x 4 to write p 2 x 4 + x 4 p x 1 p 2 x 1 = 0. This implies that x 1 = 1 + p 2 p (1 + p ) x 4 = p 1 + p 2 p (1 + p ) , x 2 = p 2 x 1 = p p 1 + p 2 1 + p From (1.2) and then (1.4), we have 1 = 1 2 x 1 2 2 = (1 + p ) 1 2 2 p 1 2 (1 + p 2 ) 1 4 , 2 = 1 2 x 4 = (1 + p ) 1 2 (1 + p 2 ) 1 4 4 p 1 2 The pair ( x, ) X R 2 + given above is the unique solution to the KuhnTucker conditions for problem ( S ). (b) We need to verify that the conditions of the KuhnTucker Sufficiency Theorem are satisfied before we can conclude that x solves problem ( S ). The set X is open and convex in R 4 . The functions f , g 1 , and g 2 are continuously differentiable on X . The function f can be expressed as the sum of functions that are concave on X , so f is concave on X . The function g 1 is linear on X , so it is concave on X . Now, the functions h 1 ( x ) = x 2 3 and h 3 ( x ) = x 2 4 are convex on X , so h 1 and h 2 are concave on X . Since we can write g 2 ( x ) = 1 h 1 ( x ) h 2 (...
View
Full
Document
 Fall '08
 MITRA
 Economics

Click to edit the document details