Question 11.4.2
The solution will follow the lecture notes style of solving KuhnTucker prob
lems:
One important point is to notice that we have to reverse on of the constraints
to be able to write the Lagrangian in the usual form. So, instead of
a
2
+
b
2
16
;
we use
±
a
2
±
b
2
² ±
16
The Lagrangian is as follows:
L
(
a;b;±
) = 2
a
2
+
b
2
±
(2
a
+
b
±
9)
±
±
(
±
a
2
±
b
2
+ 16)
Optimality Conditions:
@L
@a
= 4
a
±
2
+ 2
a±
= 0
@L
@b
= 2
b
±
+ 2
b±
= 0
(2
a
+
b
±
9) = 0
±
(
±
a
2
±
b
2
+ 16) = 0
&
0
;±
&
0
There are four possible cases:
i)
= 0
;±
= 0
4
a
= 0
2
b
= 0
a
=
b
= 0
1
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View Full DocumentFor this to be a solution it must satisfy each and every one of the optimality
conditions. We can see that the second constraint (
a
2
+
b
2
16)
Hence,
a
=
b
= 0
is not a solution.
ii)
6
= 0
;±
= 0
First order conditions become:
4
a
= 2
2
b
=
Dividing each side of both equations with each other we get
a
=
b:
Substitute this result into the constrain which we are considering as binding:
2
a
+
b
= 9
!
2
a
+
a
= 9
!
a
=
b
= 3
:
For this to be a solution one must check all optimality conditions and in fact
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 Spring '08
 cramton
 Expected utility hypothesis, order conditions, mutually exclusive events, rst order conditions

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