Problem 2.
Determine the global max and min of the function
f
(
x, y
) =
x
2

2
x
+ 2
y
2

2
y
+ 2
xy
over the compact region

1
≤
x
≤
1
,
0
≤
y
≤
2
.
Solution:
We look for the critical points in the interior:
∇
f
= (2
x

2 + 2
y,
4
y

2 + 2
x
) = (0
,
0) =
⇒
2
x

2 + 2
y
= 4
y

2 + 2
x
= 0 =
⇒
y
= 0
, x
= 1
.
However, the point
(1
,
0)
is not in the interior so we discard it for now.
We check the boundary. There are four lines to be considered:
•
the line
x
=

1
:
f
(

1
, y
) = 3 + 2
y
2

4
y.
The critical points of this function of
y
are found by setting the derivative to zero:
∂
∂y
(3 + 2
y
2

4
y
) = 0 =
⇒
4
y

4 = 0 =
⇒
y
= 1
with
f
(

1
,
1) = 1
.
•
the line
x
= 1
:
f
(1
, y
) = 2
y
2

1
.
Computing the derivative and setting it to
0
we find the critical point
y
= 0
. The corre
sponding point
(1
,
0)
is one of the corners, and we will consider it separately below.
•
the line
y
= 0
:
f
(
x,
0) =
x
2

2
x.
Computing the derivative and setting it to
0
we find
2
x

2 = 0 =
⇒
x
= 1
. This gives the
corner
(1
,
0)
as before.
•
the line
y
= 2
:
f
(
x,
2) =
x
2
+ 2
x
+ 4
with critical point
x
=

1
which is again a corner.
Finally, we check the four corners
(

1
,
0)
,
(1
,
0)
,
(

1
,
2)
,
(1
,
2)
.
The values of the function
f
are
f
(

1
,
0) = 3
,
f
(1
,
0) =

1
,
f
(

1
,
2) = 3
,
f
(1
,
2) = 7
.
From the boxed values we select the lowest and the highest to find the global min and global max.
We conclude that
global minimum occurs at
(1
,
0)
global maximum occurs at
(1
,
2)
.
Problem 3.
Using Lagrange multipliers, optimize the function
f
(
x, y
) =
x
2
+ (
y
+ 1)
2