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752
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
By Theorem 3, the extreme values occur either at a critical point in the interior of the square or at a point on the boundary
of the square. Since there are no critical points in the interior of the square, the candidates for extreme values are the
following points:
(
0
,
0
),
(
1
,
0
),
(
1
,
1
),
(
0
,
1
)
We compute
f
(
x
,
y
)
=
x
3
+
y
3
−
3
xy
at these points:
f
(
0
,
0
)
=
0
3
+
0
3
−
3
·
0
=
0
f
(
1
,
0
)
=
1
3
+
0
3
−
3
·
1
·
0
=
1
f
(
1
,
1
)
=
1
3
+
1
3
−
3
·
1
·
1
=−
1
f
(
0
,
1
)
=
0
3
+
1
3
−
3
·
0
·
1
=
1
We conclude that in the given domain, the global maximum is
f
(
1
,
0
)
=
f
(
0
,
1
)
=
1 and the global minimum is
f
(
1
,
1
)
1.
41.
f
(
x
,
y
)
=
x
3
y
5
, the set bounded by
x
=
0,
y
=
0, and
y
=
1
−
√
x
.
Hint:
Use a computer algebra system
to Fnd the minimum along the boundary curve
y
=
1
−
√
x
, which is parametrized by
(
t
,
1
−
√
t
)
for 0
≤
t
≤
1.
SOLUTION
x
y
=
1
−
x
y
A
(1, 0)
0
B
(0, 1)
We use the following steps.
Step 1.
Examine the critical points. We Fnd the critical points of
f
(
x
,
y
)
=
x
3
y
5
in the interior of the domain. Setting
the partial derivatives equal to zero and solving gives
f
x
(
x
,
y
)
=
3
x
2
y
5
=
0
,
f
y
(
x
,
y
)
=
5
y
4
x
3
=
0
The solutions are all the points with at least one zero coordinate. These points are not in the interior of the region, hence
there are no critical points in the interior of the region.
Step 2.
Check the boundary. We consider each part of the boundary separately.
The segment
OA
: On this segment
y
=
0and
f
(
x
,
0
)
=
0, hence
f
has the constant value 0 on this part of the
boundary.
The segment
OB
: On this segment
x
=
0, hence
f
(
0
,
y
)
=
0
3
·
y
5
=
0.
f
attains the constant value 0 on this part of
the boundary.
The curve
y
=
1
−
√
x
,0
≤
x
≤
1: On this curve we have
f
(
x
,
y
)
=
g
(
x
)
=
x
3
(
1
−
√
x
)
5
.
We Fnd the points on the interval 0
≤
x
≤
1where
g
(
x
)
=
x
3
(
1
−
√
x
)
5
has maximum and minimum values. Differen
tiating
g
gives
g
±
(
x
)
=
3
x
2
(
1
−
√
x
)
5
+
x
3
·
5
(
1
−
√
x
)
4
·
±
−
1
2
√
x
¶
=
3
x
2
(
1
−
√
x
)
5
−
5
2
x
2
.
5
(
1
−
√
x
)
4
=
x
2
(
1
−
√
x
)
4
±
3
−
3
√
x
−
5
2
√
x
¶
=
x
2
(
1
−
√
x
)
4
±
3
−
11
2
√
x
¶
We solve
g
±
(
x
)
=
0 in the interval 0
<
x
<
1:
g
±
(
x
)
=
x
2
(
1
−
√
x
)
4
±
3
−
11
2
√
x
¶
=
0
⇒
x
=
0
,
1
−
√
x
=
0
,
3
−
11
2
√
x
=
0
The solutions are
x
=
0,
x
=
1, and
x
=
36
121
. The critical point in the interval 0
<
x
<
1is
x
=
36
121
. We compute
g
(
x
)
=
x
3
(
1
−
√
x
)
5
at the critical point and at the endpoints
x
=
0,
x
=
1:
g
(
0
)
=
g
(
1
)
=
0
,
g
±
36
121
¶
=
±
36
121
¶
3
±
1
−
6
11
¶
5
≈
0
.
0005
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View Full Document SECTION
15.7
Optimization in Several Variables
(ET Section 14.7)
753
The points where
g
has extreme values in the interval 0
≤
x
≤
1a
re
x
=
0,
x
=
1, and
x
=
36
121
.W
efndthe
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.
 Winter '08
 GANGliu
 Critical Point

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