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Unformatted text preview: t was exactly equal to the slope of the level curve for the
objective function
,
). The feasible range: f a e g
b d c All points within the blue shaded region, and along the level set,
acceptable values for , . 1 , What is happening at the other 6 possible cases? In cases c) – e), the main constraint, ,
does not bind, so ,
. The same is true at g). At b) and f), the constraint does bind, but either 0 (case c) or 0 (case f). Only at a) will the ,
bind and , will be interior. constraint are What is happening at the other 6 possible cases? In cases c) – e), the main constraint, ,
does not bind, so ,
. The same is true at g). At b) and f), the constraint does bind, but either 0 (case c) or 0 (case f). Only at a) will the constraint ,
bind and , will be interior. Now consider our function . For a brief moment, let’s suppose is a single‐variable function, and we could graph its behavior over the non‐negative orthant.
f(x) f(x) ∗ In the first case, the optimum, given that 0
is satisfied at ∗ 0 (a corner solution
∗
∗
0) while in the second case, the solution is interior, with
0. We
with
∗
∗
could write these together as
0 and ∗
0. Given these ideas, how can we approach these three constraints of the problem we
started with (where the objective function f is a function of 2 variables)? We have
1. ∗ , ∗ ∗ ∗ , ∗ 0 and ∗ ∗ , ∗ ∗ ∗ , ∗ 0 2. ∗ , ∗ ∗ ∗ ∗ 0 and ∗ ∗ ∗ ∗ ∗ ∗ 0 3. ∗ , ∗ , 0 and ∗ ∗ , ∗ , , 0. These will handle all 7 cases of our problem (we should note, we are assuming here,
or should include, ∗ 0.
∗∗
,
. Suppose our
Consider case e) above. Here, ∗ 0, ∗ 0 and
∗∗
optimum were in this range defined by e). We see that first, b...
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 Winter '14
 Economics

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