Problem Set 4 Solution - Problem Set 4 Equality-Constrained Maximization 1 Consider the maximization problem max x1 x2 x1,x2 subject to x2 x2 = 1 1 2

# Problem Set 4 Solution - Problem Set 4 Equality-Constrained...

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Problem Set 4: Equality-Constrained Maximization 1. Consider the maximization problem max x 1 ,x 2 x 1 + x 2 subject to x 2 1 + x 2 2 = 1 Sketch the constraint set and contour lines of the objective function. Find all critical points of the Lagrangian. Verify whether each critical point is a local maximum or a local minimum. Find the global maximum. You should sketch the constraint set and contour lines/indifference curves. The constraint is a circle of radius one, and the objective is a linear hyperplane with slope 1. The Lagrangian is L = x 1 + x 2 λ ( x 2 1 + x 2 2 1) with FONCs 1 λ 2 x 1 = 0 1 λ 2 x 2 = 0 ( x 1 2 + x 2 2 1) = 0 The first two equations imply that x 1 = x 2 . The third equation implies that x 1 2 + x 2 2 = 1, or x k = ± radicalbigg 1 2 So there are two potential candidates, parenleftBigg radicalbigg 1 2 , radicalbigg 1 2 parenrightBigg , parenleftBigg radicalbigg 1 2 , radicalbigg 1 2 parenrightBigg , The Bordered Hessian is 0 2 x 1 2 x 2 2 x 1 2 λ 0 2 x 2 0 2 λ Since λ appears in it, we’ll need to compute it as well, which is different from many examples we saw in class. Also different is that the x i ’s appear along the top and left-hand borders. The determinant of the third principal minor of the Bordered Hessian is 2 x 1 (4 x 1 λ ) + 2 x 2 (4 λx 2 ) = 8 λ ( x 2 1 + x 2 2 ) So that the sign only depends on the multiplier, λ . Since λ = 1 2 x 1 = 1 2 x 2 it will be positive if both terms are positive, and negative if both terms are negative. Consequently, the critical point parenleftBigg radicalbigg 1 2 , radicalbigg 1 2 parenrightBigg 1