lecture-notes-Kuhntucker - Applications of Lagrangian Kuhn...

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Applications of Lagrangian: Kuhn Tucker Conditions Utility Maximization with a simple rationing constraint Consider a familiar problem of utility maximization with a budget constraint: Maximize U = U ( x, y ) subject to B = P x x + P y y and x > x But where a ration on x has been imposed equal to x. We now have two constraints. The Lagrange method easily allows us to set up this problem by adding the second constraint in the same manner as the fi rst. The Lagrange becomes Max x,y U ( x, y ) + λ 1 ( B P x x P y y ) + λ 2 ( x x ) However, in the case of more than one constraint, it is possible that one of the constraints is nonbinding. In the example we are using here, we know that the budget constraint will be binding but it is not clear if the ration constraint will be binding. It depends on the size of x. The two possibilities are illustrated in fi gure one. In the top graph, we see the standard utility maximization result with the solution at point E. In this case the ration constraint, x , is larger than the optimum value x . In this case the second constraint could have been ignored. In the bottom graph the ration constraint is binding. Without the constraint, the solution to the maximization problem would again be at point E. However, the solution for x violates the second constraint. Therefore the solution is determined by the intersection of the two constraints at point E’ Procedure: This type of problem requires us to vary the fi rst order conditions slightly. Cases where constraints may or not be binding are often referred to as Kuhn-Tucker conditions. The Kuhn-Tucker conditions are L x = U x P x λ 1 λ 2 = 0 x 0 L y = U y P y λ 1 = 0 y 0 and L λ 1 = B P x x P y y 0 λ 1 0 L λ 2 = x x 0 λ 2 0 Now let us interpret the Kuhn-Tucker conditions for this particular problem. Looking at the Lagrange U ( x, y ) + λ 1 ( B P x x P y y ) + λ 2 ( x x ) 1
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Figure 1: We require that λ 1 ( B P x x P y y ) = 0 therefore either λ 1 = 0 or B P x x P y y = 0 If we interpret λ 1 as the marginal utility of the budget (Income), then if the budget constraint is not met the marginal utility of additional B is zero ( λ 1 = 0 ). (2) Similarly for the ration constraint, either x x = 0 or λ 2 = 0 λ 2 can be interpreted as the marginal utility of relaxing the ration constraint. Solving by Trial and Error Solving these types of problems is a bit like detective work. Since there are more than one possible outcomes, we need to try them all. But before you start, it is important to think about the problem and try to make an educated guess as to which constraint is more likely to be nonbinding. In this example we can be sure that the budget constraint will always be binding, therefore we only need to worry about the e ff ects of the ration constraint.
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