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If x is a solution, then there exist X0 > 0 and \x such that not both are zero, and such that -2(x 1 -2)X 0 =X 1 , (i) = Xj (ii) -2(x2-2)X0 and Xj +3c2 =b. (iii) Certainly X0 =£ 0. ) Therefore, we may set X0 = 1. Then on the basis of (i) and (ii) we may conclude that X l 23 =x 2- y' A3c is a scalar. Therefore ~y'Ax = (y'Ax)' =x'A'y. 52 Notes and problems in microeconomic theory Finally, via (iii) we see that xx = 6/2, x2 = 6/2, Xx = -2(b/2 - 2) and / = -2(b/2 -If. We note that bf/db = -2(b/2 - 2) which is the same as Xi.

There are three more cases to consider. Is it possible to find a solution to (i)—(vi) with neither constraint being binding? This would imply that Xi = X2 = 0 . From (i) and (ii) we obtain X0 = 0, contradicting our assumption that X0 = 1. e. Xi > 0, X2 = 0 and *i+*2-1=0? 0*) Problem set 1 33 With X2 = 0 and X0 = 1, (i) and (ii) imply that 8 = X12x1, 9 = Xj2*2. Hence S/9=Xllx2. e. *2=±l/V(l+(8/9)2) and XX=±%I9>/(\+{%I9)2)- The negative values for {xu x2) can be ignored because they are inconsistent with a non-negative value for X2.

L l . 20 Hence the conditions of theorem 2 are fulfilled. e. cover the sign constraints by appropriate inequality conditions) to give a set of necessary and sufficient conditions for a solution. Next, consider the dual problem. 2) biyi+b2y2 subject to filial +^21^2 <*nyi ^CL22y2 >Cl, >c2 and yuy2 >0. Again, write down necessary and sufficient conditions for a solution. Discuss the relationship between the two problems. Answer. In the notation of theorem 2 we have V/=(Cl);V^=(fl»);V^=(fl21). \C2/ \012/ \022/ Hence, the non-negative vector x' = (xly x2) is a solution if and only if there exists a non-negative vector21 y' = (yl9 y2) such that x and y simultaneously satisfy 20 Actually, there is no need to worry about the constraint qualification in linear programming problems.

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An Intro to Symmetry and Supersym. in Quantum Field Theory by J. Topuszanski

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