Graphics Programs Reference
In-Depth Information
f
(
s
)
=
−
=
=
.
s
1
0
s
1
0
0
−
0
.
1
−
0
.
5
−
0
.
6
x
2
=
x
1
+
s
v
1
=
+
1
.
=
0
−
1
.
0
−
1
.
0
Wehave nowreached the minimum point.
EXAMPLE 10.6
h
b
The figure shows the cross section of a channel carrying water. Determine
h
,
b
and
that minimize the length of the wettedperimeterwhile maintaining a cross-
sectional area of 8m
2
. (Minimizing the wettedperimeterresults in least resistance to
the flow. ) Use the Fletcher-Reeves mathod.
θ
Solution
The cross-sectional area of the channel is
1
2
[
b
A
=
+
(
b
+
2
h
tan
θ
)
]
h
=
(
b
+
h
tan
θ
)
h
and the length of the wettedperimeteris
S
=
b
+
2(
h
sec
θ
)
The optimizationproblem can becast as
minimize
b
+
2
h
sec
θ
subject to (
b
+
h
tan
θ
)
h
=
8
Equality constraints can oftenbe used to eliminate some of the design variables.
In thiscase wecan solve the area constraintfor
b
,obtaining
8
h
−
b
=
h
tan
θ
Substituting the result into the expression for
S
, we get
8
h
−
S
=
h
tan
θ
+
2
h
sec
θ
Wehave now arrivedat an unconstrained optimizationproblem of finding
h
and
θ
that minimize
S
. The gradient of the meritfunctionis
h
2
∂
S
/∂
h
−
8
/
−
tan
θ
+
2 sec
θ
∇
S
=
=
h
sec
2
∂
S
/∂θ
−
θ
+
2
h
sec
θ
tan
θ
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