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
y
-coordinates oF these points From
y
=
1
−
√
x
:
x
=
0
:
y
=
1
−
√
0
=
1
x
=
1
:
y
=
1
−
√
1
=
0
x
=
36
121
:
y
=
1
−
r
36
121
=
5
11
We conclude that the global extrema oF
f
on the curve
y
=
1
−
√
x
,0
≤
x
≤
1 are obtained at the points
(
0
,
1
)
,
(
1
,
0
)
,
and
³
36
121
,
5
11
´
.
Step 3.
Conclusions. We examine the values oF
f
(
x
,
y
)
=
x
3
y
5
at the points obtained in the previous parts. The candi-
dates For global extrema are
f
=
0, the values oF
f
on the segments
OA
and
OB
:
f
(
0
,
1
)
=
f
(
1
,
0
)
=
0
f
±
36
121
,
5
11
¶
=
±
36
121
¶
3
±
5
11
¶
5
=
0
.
0005
We conclude that the minimum value oF
f
(
x
,
y
)
in the given domain is 0 and the maximum value is
f
³
36
121
,
5
11
´
≈
0
.
0005.
42.
Show that the rectangular box (including a top and bottom) with fxed volume
V
with the smallest possible surFace
area is a cube (±igure 21).
FIGURE 21
Rectangular box with sides
x
,
y
,
z
.
SOLUTION
Step 1.
±ind a Function to be maximized. The surFace area oF the box with sides lengths
x
,
y
,
z
is
S
=
2
(
xz
+
yz
+
xy
)
(1)
We express the surFace area in terms oF
x
and
y
alone using the equation
V
=
xyz
For the volume oF the box. This
equation implies that
z
=
V
, hence by (1) we get
S
=
S
(
x
,
y
)
=
2
±
x
·
V
+
y
·
V
+
¶
=
2
±
V
y
+
V
x
+
¶
=
2
V
y
+
2
V
x
+
2
That is,
S
=
2
V
y
+
2
V
x
+
2
Step 2.
Determine the domain. The variables
x
and
y
express lengths, thereFore, they must be nonnegative. Also,
S
is
not defned iF
x
=
0or
y
=
0, thereFore the domain is
D
= {
(
x
,
y
)
:
x
>
0
,
y
>
0
}