Solutions to Homework 2
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
) has a relative maximum at (1
,

1).
Since
D
(0
,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Sarason
 Equations, Derivative, Optimization, 8 feet, 12 feet, constraint equation

Click to edit the document details