14.7 # 8 (10 points):
Find the critical points of
f
(
x,y
) =
x
3
+ 2
xy

2
y
2

10
x
. Use
the Second Derivative Test to determine whether they are local minima or maxima.
Solution:
The critical points of
f
are the points where
∇
f
is not deﬁned, or is zero.
Since
f
is a polynomial, then
f
x
and
f
y
exists for all (
x,y
) in the domain of
f
. So we
need to solve
∇
f
=
h
3
x
2
+ 2
y

10
,
2
x

4
y
i
=
h
0
,
0
i
.
Equating the second coordinate gives
y
=
1
2
x
. Substituting into the equation from the
ﬁrst coordinate gives
3
x
2
+ 2(
1
2
x
)

10 = 0
(
x
+ 2)(3
x

5) = 0
,
so
x
=

2 or
x
=
5
3
. Since
y
=
1
2
x
, then the critical points of
f
are
(

2
,

1) and (
5
3
,
5
6
)
.
Next, calculate the discriminant.
D
(
x,y
) = det
±
6
x
2
2

4
²
=

24
x

4
.
D
(

2
,

1) = 44
>
0, and
f
xx
(

2
,

1) =

12
<
0, so
f
has a local maximum at
(

2
,

1), (and the value is
f
(

2
,

1) = 14).
D
(
5
3
,
5
6
) =

44
<
0, so
f
has a saddle point at (
5
3
,
5
6
).
14.8 # 16 (10 points):
Find the rectangular box of maximum volume if the sum of
the lengths of the edges is
300
cm.
Solution
: Let
x,y,z
denote the width, length and height of the box. We want to
maximize the volume
V
(
x,y,z
) =
xyz,
subject to the constraint
g
(
x,y,z
) = 4(
x
+
y
+
z
) = 300
.
Use the Lagrange multiplier method and solve
∇
V
=
λ
∇
g
and
g
= 300 simultaneously.
Since