Fall 2017
Discussion Set (Week 7)
Solutions
Calculus II
2. Let
f
(
x, y
) =
x
3
+
y
3
+ 3
x
2

3
y
2

8.
(a) Find
f
x
and
f
y
.
Answer:
f
x
= 3
x
2
+ 6
x
=
3
x
(
x
+ 2)
and
f
y
= 3
y
2

6
y
=
3
y
(
y

2)
.
(b) Find the critical points of
f
. What does the
First Derivative Test
tell you about these points?
Answer:
f
x
and
f
y
exist everywhere, so the critical points are those at which both
f
x
and
f
y
vanish, i.e.,
(0
,
0)
,
(0
,
2)
,
(

2
,
0)
, and
(

2
,
2)
. The First Derivative Test says that
if
f
has a
local extremum, then it must occur at one of these points.
(c) Find the
Hessian
H
f
= det
f
xx
f
xy
f
yx
f
yy
.
Answer:
Since
f
xx
= 6
x
+ 6
,
f
xy
= 0, and
f
yy
= 6
y

6, the Hessian is
H
f
= det
f
xx
f
xy
f
yx
f
yy
= det
6
x
+ 6
0
0
6
y

6
= (6
x
+ 6)(6
y

6) =
36(
x
+ 1)(
y

1)
.
(d) By applying the
Second Derivative Test
,
i. find the local minima of
f
and their locations
ii. find the local maxima of
f
and their locations.
iii. find the saddle points of
f
and their locations.
Answer:
We first look at the sign of the Hessian at the critical points:
H
f
(0
,
0)
<
0
,
H
f
(0
,
2)
>
0
,
H
f
(

2
,
0)
>
0
,
H
f
(

2
,
2)
<
0
.
Next, we look at the sign of
f
xx
at the critical points:
f
xx
(0
,
0)
>
0
,
f
xx
(0
,
2)
>
0
,
f
xx
(

2
,
0)
<
0
,
f
xx
(

2
,
2)
<
0
.
By the Second Derivative Test,
f
has a local minimum at
(0
,
2)
, a local maximum at
(

2
,
0)
,
and saddle points at
(0
,
0)
and
(

2
,
2)
. The local minimum value is
f
(0
,
2) =

12
, the local
maximum value is
f
(

2
,
0) =

4
, and the saddle points are (0
,
0
, f
(0
,
0
,
0)) =
(0
,
0
,

8)
and
(

2
,
2
, f
(

2
,
2)) =
(

2
,
2
,

8)
.
3. For each of the following functions defined on
R
2
, find the critical points, local extrema, and saddle
points, if any. How useful are the
First Derivative Test
and the
Second Derivative Test
in finding
these points?
(a)
f
(
x, y
) =
p
x
2
+
y
2
Answer:
f
(
x,
0) =

x

is not differentiable at
x
= 0, so
f
x
does not exist at (0
,
0). (Neither
does
f
y
.)
Hence, (0
,
0) is a critical point of
f
.
On the other hand, if (
x, y
)
6
= (0
,
0), then
f
x
=
x
√
x
2
+
y
2
and
f
y
=
y
√
x
2
+
y
2
, so
f
x
and
f
y
exist on
R
2
 {
(0
,
0)
}
, and they cannot be both
zero.
It follows that
(0
,
0)
is the only critical point of
f
.
It is clear that
f
(0
,
0) =
0
is an
absolute minimum, and the First Derivative Test implies that it is the only local extremum.