PS6Solutions

PS6Solutions - Question 11.4.2 The solution will follow the...

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Question 11.4.2 The solution will follow the lecture notes style of solving Kuhn-Tucker prob- lems: One important point is to notice that we have to reverse on of the constraints to be able to write the Lagrangian in the usual form. So, instead of a 2 + b 2 16 ; we use ± a 2 ± b 2 ² ± 16 The Lagrangian is as follows: L ( a;b;± ) = 2 a 2 + b 2 ± (2 a + b ± 9) ± ± ( ± a 2 ± b 2 + 16) Optimality Conditions: @L @a = 4 a ± 2 + 2 = 0 @L @b = 2 b ± + 2 = 0 (2 a + b ± 9) = 0 ± ( ± a 2 ± b 2 + 16) = 0 & 0 & 0 There are four possible cases: i) = 0 = 0 4 a = 0 2 b = 0 a = b = 0 1
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For this to be a solution it must satisfy each and every one of the optimality conditions. We can see that the second constraint ( a 2 + b 2 16) Hence, a = b = 0 is not a solution. ii) 6 = 0 = 0 First order conditions become: 4 a = 2 2 b = Dividing each side of both equations with each other we get a = b: Substitute this result into the constrain which we are considering as binding: 2 a + b = 9 ! 2 a + a = 9 ! a = b = 3 : For this to be a solution one must check all optimality conditions and in fact
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PS6Solutions - Question 11.4.2 The solution will follow the...

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