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
13 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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 Winter '14
 Economics, Derivative, producer, objective function, Constraint, Gradient, quasiconcavity

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