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Λw ν note that neither c 0 or r 0 allows a solution

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λw + ν Note that neither c = 0 or r = 0 allows a solution. Thus μ = ν = 0 by complementary slackness. Moreover, r = T implies c = 0, so it is also impossible. Thus ρ = 0 by complementary slackness. However, λ = 1 /pc > 0, so pc + rw = wT is the only binding constraint. Constraint qualification is clearly satisfied with this one constraint. We solve the first-order equa- tions, obtaining c = 2 wT 3 p and r = T 3 . Finally, we need only look at the determinant of the bordered Hessian because there is one constraint and two unknowns. The bordered Hessian is H = 0 p w p - c - 2 0 w 0 - 2 r - 2 . Its determinant is 2( p/r ) 2 +4( cw ) 2 > 0. Since n = 2, det H ( - 1) 2 > 0, which implies we have a maximum. 5. Let A be an n × n positive definite matrix. Define N ( x ) = x T Ax . Is N a norm? That is, does it obey the three conditions a norm must obey? Answer: It is a norm. 1) It is absolutely homogeneous of degree 1 by construction. 2) It is clearly non-negative, and if N ( x ) = 0, the quadratic form x T Ax = 0. Since this form is positive definite, x = 0. Finally, the triangle inequality must be shown.
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