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# slide14-sept23 - ECON 401: Theorem of Khun and Tucker...

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ECON 401: Theorem of Khun and Tucker Siyang Xiong Rice University September 20, 2011 Xiong (Rice University) ECON 401 September 20, 2011 1 / 21

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An example x . & 0 / to produce two outputs y . & 0 / , z . & 0 / by the following technology: y 2 C z 2 ± x . The price for x , y , z are 1, 2, 3 respectively. What is the optimal production plan? Xiong (Rice University) ECON 401 September 20, 2011 2 / 21
An example max . 2 y C 3 z x / s.t. x y 2 z 2 ± 0 I x ± 0 I y ± 0 I z ± 0 . Step 1: write down the Lagrange: L D . 2 y C 3 z x / C 1 x y 2 z 2 ± C 2 x C 3 y C 4 z Xiong (Rice University) ECON 401 September 20, 2011 3 / 21

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An example Step 2: take derivatives @ L @ x 1 C 1 C 2 D 0 (1) @ L @ y D 2 2 1 y C 3 D 0 (2) @ L @ z D 3 2 1 z C 4 D 0 (3) x y 2 z 2 ± 0 I 1 ± 0 I 1 x y 2 z 2 ± D 0 . (4) x ± 0 I 2 ± 0 I 2 x D 0 . (5) y ± 0 I 3 ± 0 I 3 y D 0 . (6) z ± 0 I 4 ± 0 I 4 z D 0 . (7) Xiong (Rice University) ECON 401 September 20, 2011 4 / 21
Step 3: solve the system of equations derived at Step 2: by (2), 1 6D 0 and y 6D 0, because if not, then 3 2 hence, by (4) and (6), x y 2 z 2 ± D 0 (8) 3 D 0 similarly, by (3), z 6D 0, because if not, then 4 3 hence, by (7), 4 D 0 Also, we cannot have 2 > 0, because if not, then by (5), we have x D 0. Then, (8), i.e., ² x y 2 z 2 ³ D 0, implies y D z D 0, which contradicts to y 6D 0 and z 6D 0. Therefore,

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## This note was uploaded on 02/09/2012 for the course ECON 401 taught by Professor Siyang during the Spring '11 term at Rice.

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slide14-sept23 - ECON 401: Theorem of Khun and Tucker...

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