7
.
3
.
2 First we take the partial derivatives and set them equal to 0 to obtain.
f
x
=
x

3 = 0 and
f
y
= 2
y
+ 2 = 0
.
Solving these equations the only possible point where a relative extremum
can exist is (3
,

2).
7
.
3
.
8 First we take the partial derivatives and set them equal to 0 to obtain.
f
x
= 4
x
3

8
y
= 0 and
f
y
=

8
x
+ 4
y
= 0
.
Solving both of these equations for
y
we obtain
y
=
x
3
/
2 and
y
= 2
x.
Setting these equal to one another we obtain
x
3
2
= 2
x
x
3

4
x
= 0
x
(
x

2)(
x
+ 2) = 0
Solving for
x
and plugging back in for
y
we see the only possible points where
a relative extremum can exist are (

2
,

4), (0
,
0), and (2
,
4).
7
.
3
.
11 First we compute the second derivatives:
f
xx
= 6,
f
xy
=

6, and
f
yy
= 6
y
. Then we deﬁne
D
(
x,y
) = 36
y

36. Since
D
(3
,
3) = 72
>
0 and
f
xx
(3
,
3) = 6
>
0 we know that
f
(
x,y
) has a relative minimum at (3
,
3).
Since
D
(

1
,

1) =

72
<
0 we know that
f
(
x,y
) has a saddle point at
(

1
,

1).
7
.
3
.
12 First we compute the second derivatives:
f
xx
=

12
x
,
f
xy
= 12
x
,
and
f
yy
= 12

36
y
2
. Hence we have
D
(
x,y
) = 432
xy
2

144
x

144
x
2
.
Since
D
(1
,
1) = 144
>
0 and
f
xx
(1
,
1) =

12
<
0 we know that
f
(
x,y
) has
a relative maximum at (1
,
1). Since
D
(1
,

1) = 144
>
0 and
f
xx
(1
,

1) =

12
<
0 we know that
f
(
x,y
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 Spring '06
 Sarason
 Equations, Derivative

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