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
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is a maximum while parenleftBigg radicalbigg 1 2 , radicalbigg 1 2 parenrightBigg is actually a minimum. Since the two critical points where x and y have the same sign are local maxima and yield the same value of the objective function, they are both global maxima. 2. Consider the maximization problem max x,y xy subject to a 2 x 2 + b 2 y 2 = r 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. How does the value of the objective vary with r ? a ? How does x respond to a change in r ; show this using the closed-form solutions and the IFT. Sketch. The constraint set is an ellipse and the objective function is “Cobb-Douglas”. The Lagrangian is L ( x, λ ) = xy λ ( a 2 x 2 + b 2 y 2 r ) with FONCs y λax = 0 x λby = 0 ( a 2 x 2 + b 2 y 2 r ) = 0 The first two equations imply that y/x = ax/ ( by ), or by 2 = ax 2 . Using the third equation, we get ax 2 = r or x = ± radicalbigg r a implying that y = ± radicalbigg r b so we have four critical points, parenleftbiggradicalbigg r a , radicalbigg r b parenrightbigg , parenleftbiggradicalbigg r a , radicalbigg r b parenrightbigg , parenleftbigg radicalbigg r a , radicalbigg r b parenrightbigg , parenleftbigg radicalbigg r a , radicalbigg r b parenrightbigg ,
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