Math 192, Prelim 2
April 14, 2005
Solutions
1) a) Consider
f
(
x, y
) =

x
2

y
2
. Then
f
x
=

2
x
and
f
y
=

2
y
so the only critical point
is (0
,
0). Observe
f
xx
(0
,
0) =

2 =
f
yy
(0
,
0) and
f
xy
(0
,
0) =
f
yx
(0
,
0) = 0 so
D
(0
,
0) = 4
>
0
and we have a maxima.
b) Consider
f
(
x, y
) =
x
2

y
2
. Then
f
x
= 2
x
and
f
y
=

2
y
so the only critical point is (0
,
0).
Observe
f
xx
(0
,
0) = 2,
f
yy
(0
,
0) =

2 and
f
xy
(0
,
0) =
f
yx
(0
,
0) = 0 so
D
(0
,
0) =

4
<
0 and
we have a saddle.
2) Since
f
(
x, y
) =
x
2
+
xy
we see
f
x
= 2
x
+
y
and
f
y
=
x
.
Solving 2
x
+
y
= 0 and
x
= 0 simultaneously gives that (0
,
0) is the only critical point. Note (0
,
0) is in the region
consisting of points (
x, y
) satisfying
x
2
+
xy
+
y
2
≤
1. As
f
xx
(0
,
0) = 2,
f
yy
(0
,
0) = 0 and
f
xy
(0
,
0) =
f
yx
(0
,
0) = 1 we see
D
(0
,
0) =

1
<
0 so (0
,
0) is a saddle.
We now analyze the boundary given by
g
(
x, y
) =
x
2
+
xy
+
y
2

1 = 0 using Lagrange
multipliers. Setting
~
∇
f
=
λ
~
∇
g
we see 2
x
+
y
=
λ
(2
x
+
y
) and
x
=
λ
(2
y
+
x
). From the first
equation we see either
λ
= 1 or 2
x
+
y
= 0.
If
λ
= 1 the second equation gives
y
= 0. Plugging this back into the constraint equation
g
we see
x
=
±
1. Note
f
(1
,
0) =
f
(

1
,
0) = 1.