which Slater’s Condition is violated, and we can’t
f
nd a saddle point. Try to
understand the di
f
erence between that example and this problem.
2. [Application of the theorem: “GM
⇒
SP”]
Let
a
∈
(0
,
1)
and
b
∈
(0
,
1)
be given parameters. Let
A
be the set de
f
ned
by:
A
=
{
(
x, y, z
)
∈
R
3
+
:
y
≤
x, z
≥
bx
and
[
y
+
a
(
z
−
bx
)]
≤
1
}
Consider the following constrained optimization problem:
Maximize y
subject to
(
x, y, z
)
∈
A
and
z
≥
x
⎫
⎬
⎭
(
B
)
Problem
(
B
)
has a solution (you do not have to show this); denote an arbitrary
solution of problem
(
B
)
by
(¯
x,
¯
y,
¯
z
)
.
Show that there is
¯
p>
0
such that:
y
+¯
pz
−
¯
px
≤
¯
y
for all
(
x, y, z
)
∈
A
Proof:
The title of the problem serves as a hint: use Theorem 45 in the Lecture
Notes. It is given that
x,
¯
¯
z
)
is a constrained global maximum. Let
g
(
x, y, z
)=
z
−
x, f
(
x, y, z
y, X
=
A,
we need to check
f
rst that
A
is convex,
f,g
are
concave. All of these can be done by checking the de
f
nitions of convex sets and
concave functions. Is the Slater’s Condition satis
f
ed? Yes, since
(0
,
0
,
1)
in an
element in
A,
and
g
(0
,
0
,
1) = 1
>
0
.