Lecture 7

# The same can be said about move along if they

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Unformatted text preview: we knew, since the constraint binds. ∗ b) We notice that for 2) ∗ , ∗ ∗ ∗ , ∗ ∗ , ∗ ∗, ∗ ∗, ∗ ∗ , ∗ 0 Then, 0. This condition, that , , , , You could think of the term ∗, ∗ ∗, ∗ function from a marginal change in the constraint , , should make intuitive sense in this case. , loosely, as the marginal increment in the objective ∗ (adjusted for the fact that changes in ). The same can be said about ∗, ∗ ∗, ∗ ∗ move along . If they are equal, then, at the marginal they sort of give equal ‘push’ to the objective function , . This is what happens in Case a). If one variable gives greater ‘push’ than the other at the marginal, we’d expect the other variable isn’t a factor, e.g., ∗ 0. The above conditions 1‐3 are referred to as Kuhn‐Tucker conditions. It is important to note that these are necessary but not sufficient conditions for an optimum. However, if we have enough conditions to ensure the solution to these conditions is unique, then we are assured that the solution to the Kuhn‐Tucker conditions will be the solution to our maximization problem. Taking all this into account, we have a general formulation for our first‐order conditions and the K‐T conditions: , ≡ Here, . Note we are using to denote a constraint i, it isn’t referring to the partial derivative of with respect to . We call this a Kuhn‐Tucker Lagrangian. We have left out the non‐negatively constraints for each but is implicit in our formulation. Our first order conditions are then: , 1. 2. , 3. 0 for 4. 0 for 0 for 0 for 1,2, … 1,2, … 1,2, … . 1,2, … . An Example Consider the problem of a firm that produces two product lines, and ....
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## This document was uploaded on 02/07/2014.

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