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Lecture 7

# Here because we know that the constraint does not bind

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Unformatted text preview: ecause , , ∗ relaxing the constraint will not increase the values of , . Because 0, it ∗∗ , 0. Finally, any marginal change in ∗ cannot raise , , must be that ∗∗ ∗∗ i.e., , ≯ 0, , 0, or otherwise, we would not set ∗ 0. ∗∗ , is inside the inferior set of Likewise, consider case g). Here, because ∗∗ ∗∗ , , we know that the constraint , does not bind, i.e., ∗∗ , . Hence, ∗ 0. Additionally, we see that ∗ 0 and ∗ 0, meaning ∗∗ ∗∗ that , 0 and , 0. These conditions all follow from our conditions 1-3 above. Last, consider case b). This would look like: 2 1 b What is happening at this point? First, the constraint ∗∗ , . ∗ , ∗ 0 is binding, so Now, the fact that the solution at b) is characterized by ∗ 0 and where ,0 , will only increase if we move outside of the shaded blue area means that defined by the constraint , 0. At b), then, the slope of the level set defined by the constraint , 0 is not as steep as the level set defined by , , where is some number, as we see in the graph. We can determine the slope of the level curve for the constraint, , by implicitly solving for in terms of . (We’ve done this before). Along the level set, we have , , , Likewise, along the level curve defined by , , , , Our decision leads us to: Rearranging, we have: , , , , , , . , we have . . , or , , , , . Now, let’s go back to the solution and try to make sense of all this… We have: a) ∗ 0. This implies that as we increase ∗ our objective function first‐order condition should hold with equality (note that here, since the constraint binds at b), a marginal change in ∗ subject to moving along the constraint , implies that ∗ , ∗ ∗ ∗ , ∗ 0 or ∗, ∗ ∗ ∗, ∗ 0, which...
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