Exam 2 Solution 2014 on Mathematical Methods of Economic Analysis

There are three variables n 3 and one linear

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There are three variables ( n = 3) and one linear constraint ( m = 1), so we look at the last n - m = 2 leading principal minors. They are H 3 = - 1 and H 4 = +1 / 4. Since H 4 ( - 1) n = H 4 ( - 1) m = - 1 / 4 < 0, the quadratic form fails both the tests for positive definiteness and negative definiteness on the constraint set. As H 4 is non-zero, we may conclude that H is indefinite on the constraint set and that (0 , 0 , 0) is a saddlepoint. Alternatively, consider the points x 1 = ε (1 , 2 , - 2) and x 2 = ε (1 , - 2 , 2). Both satisfiy the constraint and Q ( x 1 ) = - ε 2 and Q ( x 2 ) = 3 ε 2 . This shows that Q takes both positive and negative values in any neighborhood of 0 , so 0 is neither a constrained local max, nor constrained local min.
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