Volume 04 Issue 06-2024
57
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
57-65
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
A
BSTRACT
The article discusses conditional programming problems. Such problems can in principle, be solved using
classical methods. However, along this path there are computational difficulties that make it necessary to
search for other solution methods. Therefore, in this article we proposed particular methods for solving
nonlinear programming problems.
K
EYWORDS
Unconditional and conditional problems, extremum problems, greatest and least values of the product.
I
NTRODUCTION
If in a problem for an extremum
𝑓(𝑥
1
, 𝑥
2
, … 𝑥
𝑛
) → 𝑚𝑎𝑥
(1)
Journal
Website:
http://sciencebring.co
m/index.php/ijasr
Copyright:
Original
content from this work
may be used under the
terms of the creative
commons
attributes
4.0 licence.
Research Article
METHODS FOR SOLVING UN CONDITIONAL AND
CONDITIONAL EXTREMUM PROBLEMS
Submission Date:
June 20,
2024,
Accepted Date:
June 25, 2024,
Published Date:
June 30, 2024
Crossref doi:
https://doi.org/10.37547/ijasr-04-06-11
Saipnazarov Shaylovbek Aktamovich
Associate Professor, Candidate of Pedagogical Sciences, Uzbekistan
Khodjabaeva Dilbar
Senior Lecturer, Tashkent University of Economics, Uzbekistan
Ortiqova Malika
Senior Lecturer, Tashkent University of Economics, Uzbekistan
Volume 04 Issue 06-2024
58
International Journal of Advance Scientific Research
(ISSN
–
2750-1396)
VOLUME
04
ISSUE
06
Pages:
57-65
SJIF
I
MPACT
FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
there are no restrictions on variables, then such a
problem is called an unconditional problem for an
extremum. The following problem for the
extremum.
𝑓(𝑥) → 𝑚𝑖𝑛
(2)
𝑔
𝑖
(𝑥) ≥ 0; 𝑖 = 1,2, … , 𝑚, 𝑥 ∈ 𝑅
𝑛
is called the conditional minimum problem of
nonlinear programming.
Example - 1.
Find all pairs
(𝑥, 𝑦)
of positive
numbers at which the smallest value of the
function is achieved
𝑓(𝑥, 𝑦) =
𝑥
4
𝑦
4
+
𝑦
4
𝑥
4
−
𝑥
2
𝑦
2
−
𝑥
2
𝑦
2
+
𝑥
𝑦
+
𝑦
𝑥
Solution.
Since there are relations
𝑓(𝑥, 𝑦) − 2 = (
𝑥
2
𝑦
2
− 1)
2
+ (
𝑦
2
𝑥
2
− 1)
2
+ (
𝑥
𝑦
−
𝑦
𝑥
)
2
+ (
𝑥
𝑦
− 2 +
𝑦
𝑥
) ≥
(𝑥 − 𝑦)
2
𝑥𝑦
≥ 0
and equality
𝑓(𝑥, 𝑦) = 2
is achieved if and only if
𝑥 = 𝑦
.
Example - 2.
If for positive numbers
𝑎, 𝑏, 𝑐
and
𝑎𝑏𝑐 = 1
, then find the minimum value expression
those
1
𝑎
3
(𝑏 + 𝑐)
+
1
𝑏
3
(𝑐 + 𝑎)
+
1
𝑐
3
(𝑎 + 𝑏)
→ 𝑚𝑖𝑛
Solution.
Convenient to move to new variables
𝑥 =
1
𝑎
, 𝑦 =
1
𝑏
, 𝑧 =
1
𝑐
,
also positive and related by
condition
𝑥𝑦𝑧 = 1
. This expression is equivalent
to the following:
𝑆 =
𝑥
2
𝑦 + 𝑧
+
𝑦
2
𝑧 + 𝑥
+
𝑧
2
𝑥 + 𝑦
→ 𝑚𝑖𝑛
Applying the Cauchy-Bunyakovsky inequality to
vectors
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(ISSN
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VOLUME
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Pages:
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(2023:
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(2024:
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OCLC
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𝑢̅ =
𝑥
√𝑦+𝑧
+
𝑦
√𝑧+𝑥
+
𝑧
√𝑥+𝑦
and equality
𝑣̅ = (√𝑦 + 𝑧, √𝑧 + 𝑥, √𝑥 + 𝑦)
we obtain
𝑢̅𝑣̅ ≤ |𝑢̅||𝑣̅|
(𝑥 + 𝑦 + 𝑧)
2
≤ 2𝑆(𝑥 + 𝑦 + 𝑧)
, those
𝑆 ≥
𝑥+𝑦+𝑧
2
Using the inequality between the geometric
mean of three positive numbers we get:
𝑆 ≥
1
2
(𝑥 + 𝑦 + 𝑧) ≥
3
2
√𝑥𝑦𝑧
3
=
3
2
Example - 3.
For a given number
𝑛 ≥ 2,
find the
largest and smallest values of
𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
provided that
𝑥
𝑖
≥
1
𝑛
(𝑖 = 1,2, … , 𝑛)
and
𝑥
1
2
+ 𝑥
2
2
+ ⋯ + 𝑥
𝑛
2
= 1
Solution.
1) Let’s find the smallest value of the
product
𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
. Let an arbitrary set
(𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
)
satisfy the conditional problem.
Consider
a
new
set
(𝑦
1
, 𝑦
2
, … , 𝑦
𝑛
)
where
𝑦
1
= 𝑥
1
, 𝑦
2
= 𝑥
2
, … , 𝑦
𝑛−2
= 𝑥
𝑛−2
𝑦
𝑛−1
= √𝑥
𝑛−1
2
+ 𝑥
𝑛
2
−
1
𝑛
2
,
𝑦
𝑛
=
1
𝑛
This set satisfies the relations
𝑦
𝑖
≥
1
𝑛
, (𝑖 = 1,2, … , 𝑛), ∑
(𝑦
𝑖
)
2
= 1
𝑛
𝑖=1
Let’s prove
Volume 04 Issue 06-2024
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International Journal of Advance Scientific Research
(ISSN
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2750-1396)
VOLUME
04
ISSUE
06
Pages:
57-65
SJIF
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FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
𝑥
1
𝑥
2
… 𝑥
𝑛
≥ 𝑦
1
𝑦
2
… 𝑦
𝑛
Indeed, we have
𝑥
𝑛−1
2
∙ 𝑥
𝑛
2
− (𝑦
𝑛−1
𝑦
𝑛
)
2
= 𝑥
𝑛−1
2
∙ 𝑥
𝑛
2
− (𝑥
𝑛−1
2
+ 𝑥
𝑛
2
−
1
𝑛
2
)
2
∙
1
𝑛
=
= (𝑥
𝑛−1
2
−
1
𝑛
2
) (𝑥
𝑛
2
−
1
𝑛
2
) ≥ 0
Further, let’s put
𝑦
1
(1)
= 𝑦
1
, … , 𝑦
𝑛−3
(1)
= 𝑦
𝑛−3
𝑦
𝑛−2
(1)
= √𝑦
𝑛−2
2
+ 𝑦
𝑛−1
2
−
1
𝑛
2
, 𝑦
𝑛−1
(1)
= 𝑦
𝑛
(1)
=
1
𝑛
and similarly we get that
𝑦
𝑖
(1)
≥
1
𝑛
,
∑
(𝑦
𝑖
(1)
)
2
= 1
𝑛
𝑖=1
and
𝑦
1
𝑦
2
… 𝑦
𝑛
≥ 𝑦
1
(1)
𝑦
2
(1)
… 𝑦
𝑛
(1)
Repeating this procedure
(𝑛 − 1)
times, we will
eventually get the set
(𝑦
1
(𝑛−1)
𝑦
2
(𝑛−1)
… 𝑦
𝑛
(𝑛−1)
)
where
𝑦
1
(𝑛−1)
= √
𝑛
2
− 𝑛 + 1
𝑛
,
𝑦
2
(𝑛−1)
= ⋯ = 𝑦
𝑛
(𝑛−1)
=
1
𝑛
Volume 04 Issue 06-2024
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(ISSN
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VOLUME
04
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06
Pages:
57-65
SJIF
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FACTOR
(2022:
5.636
)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
and
𝑦
𝑖
(𝑛−1)
≥
1
𝑛
,
∑
(𝑦
𝑖
(𝑛−1)
)
2
= 1
𝑛
𝑖=1
This means that for any set
(𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
),
satisfying the conditional problem, the inequality
is true
𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
≥ √
𝑛
2
− 𝑛 + 1
𝑛
𝑛
and at
𝑥
1
= √
𝑛
2
− 𝑛 + 1
𝑛
, 𝑥
2
=, … , 𝑥
𝑛
=
1
𝑛
equality is achieved. So the smallest value is
√
𝑛
2
− 𝑛 + 1
𝑛
𝑛
2) Let’s find the greatest value of the product
𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
. Applying of the theorem on averages,
we get
𝑥
1
2
𝑥
2
2
… 𝑥
𝑛
2
≤ ((
𝑥
1
2
+ 𝑥
2
2
+ ⋯ + 𝑥
𝑛
2
𝑛
))
𝑛
=
1
𝑛
𝑛
,
those
𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
≤ 𝑛
−
𝑛
2
Volume 04 Issue 06-2024
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(ISSN
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VOLUME
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Pages:
57-65
SJIF
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(2022:
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(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
Equality is achieved at
𝑥
1
= 𝑥
2
= ⋯ = 𝑥
𝑛
=
1
√𝑛
So the largest value is
𝑛
−
𝑛
2
.
Example - 4.
For given numbers
𝑛 ∈ 𝑁,
and
𝑎 ∈
[1; 𝑛],
find the largest value of the expression
|∑
𝑠𝑖𝑛2𝑥
𝑖
𝑛
𝑖=1
|
provided that
∑
𝑠𝑖𝑛2𝑥
𝑖
= 𝑎
𝑛
𝑖=1
Solution.
We have
𝑎 = ∑
𝑠𝑖𝑛2𝑥
𝑖
=
𝑛
𝑖=1
∑
1 − 𝑐𝑜𝑠2𝑥
𝑖
2
,
𝑛
𝑖=1
∑
𝑐𝑜𝑠2𝑥
𝑖
= 𝑛 − 2𝑎
𝑛
𝑖=1
Next, consider vectors
(𝑐𝑜𝑠2𝑥
𝑖
, 𝑠𝑖𝑛2𝑥
𝑖
)
Of unit length on the plane. Their sum has length
no more than
𝑛
, which means the inequality holds
|∑
𝑠𝑖𝑛2𝑥
𝑖
𝑛
𝑖=1
| ≤ √𝑛
2
− (∑
𝑐𝑜𝑠2𝑥
𝑖
𝑛
𝑖=1
)
2
= √𝑛
2
− (𝑛 − 2𝑎)
2
= 2√𝑎(𝑛 − 𝑎),
is satisfied, in which equality is
achieved at
𝑥
1
= 𝑥
2
= ⋯ = 𝑥
𝑛
= 𝑎𝑟𝑐𝑠𝑖𝑛√
𝑎
𝑛
. So the largest value is
2√𝑎(𝑛 − 𝑎).
It is easy to see that the problem of determining a
conditional extremum coincides with the
problem of nonlinear programming.
One way to determine conditional extremum is
used if
"𝑚"
variables from the relationship
equations, for example
𝑥
1
= 𝑥
2
= ⋯ = 𝑥
𝑚
,
can be
explicity expressed in terms of the remaining
"𝑛 − 𝑚"
variables:
𝑥
𝑖
= 𝑔
𝑖
(𝑥
𝑚+1
, … , 𝑥
𝑚
), 𝑖 = 1,2, … , 𝑚
Substituting the resulting expressions for
𝑥
𝑖
into
the function
𝑍
we obtain
𝑍 = 𝑓(𝑔(𝑥
𝑚+1
, … , 𝑥
𝑚
), … , 𝑔
𝑚
(𝑥
𝑚+1
, … , 𝑥
𝑚
), 𝑥
𝑚+1
, … , 𝑥
𝑚
),
or
Volume 04 Issue 06-2024
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VOLUME
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Pages:
57-65
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(2023:
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)
(2024:
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)
OCLC
–
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𝑍 = 𝐹(𝑥
𝑚+1
, … , 𝑥
𝑚
).
The problem is reduced to finding a local (global)
extremum for a function of
"𝑛 − 𝑚"
variables.
Example - 5.
The flour mill sells flour in two
ways: retail through a store and wholesale
through sales agents. When selling
𝑥
1
kg of flour
through a store, sales costs are
𝑥
1
2
rubles, and
when selling
𝑥
2
kg of flour through sales agents
are
𝑥
2
2
rubles. Determine how many kilograms of
flour should be sold in lack way so that sales costs
are minimal if 5000 are allocated for sale per day
kg flour.
Solution.
Let’s create a mathematical model of
the problem. Let’s find the minimum total costs
𝐹(𝑥) = 𝑥
1
2
+ 𝑥
2
2
(3)
with restrictions
𝑥
1
+ 𝑥
2
= 5000, 𝑥
1
≥ 0, 𝑥
2
≥ 0 (4)
It is necessary to find the conditional extremum
of function (3), if the connection equation has the
form (4). From the equation we find, for example,
𝑥
2
,and substitute it in (3).
𝑥
2
= 5000 − 𝑥
1
, 𝐹 = 𝑥
1
2
+ (5000 − 𝑥
1
)
2
(5)
Wherein
𝑥
1
∈ [0; 5000]
. Let’s find th
e
global extremum of function (5) on the interval
[0; 5000]
.
The stationary point is equal to 2500, from
the definition of the function we obtain that
𝐹
at
𝑥
1
= 𝑥
2
= 2500
reaches a minimum.
Example - 6.
Find the greatest value of the
product
𝑥
1
2
𝑥
2
2
𝑥
3
2
𝑥
4
,
provided that given number
𝑛 ≥ 2,
find the largest and smallest values of
𝑥
1
, 𝑥
2
, 𝑥
3
, 𝑥
4
≥ 0
and
2𝑥
1
+ 𝑥
1
∙ 𝑥
2
+ 𝑥
3
+ 𝑥
2
∙ 𝑥
3
∙ 𝑥
4
= 1
Volume 04 Issue 06-2024
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(ISSN
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VOLUME
04
ISSUE
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Pages:
57-65
SJIF
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(2022:
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)
(2023:
6.741
)
(2024:
7.874
)
OCLC
–
1368736135
Solution.
Using the mean theorem, we have
√2𝑥
1
2
𝑥
2
2
𝑥
3
2
𝑥
4
4
= √2𝑥
1
∙ 𝑥
1
𝑥
2
∙ 𝑥
3
∙ 𝑥
2
𝑥
3
𝑥
4
4
≤
2𝑥
1
+ 𝑥
1
∙ 𝑥
2
+ 𝑥
3
+ 𝑥
2
∙ 𝑥
3
∙ 𝑥
4
4
=
1
4
those
𝑥
1
2
𝑥
2
2
𝑥
3
2
𝑥
4
≤
1
512
Equality is achieved if
2𝑥
1
= 𝑥
1
∙ 𝑥
2
= 𝑥
3
= 𝑥
2
∙ 𝑥
3
∙ 𝑥
4
=
1
4
those at
𝑥
1
=
1
8
, 𝑥
2
= 2, 𝑥
3
=
1
4
, 𝑥
4
=
1
2
.
So the largest value is
1
512
.
Example - 7.
For given numbers
𝑛 ≥ 2
and
𝑎 >
0,
find the largest value of the sum
∑
𝑥
𝑖
𝑛−1
𝑖=1
𝑥
𝑖+1
provided that
𝑥
𝑖
≥ 0 (𝑖 = 1,2, … , 𝑛)
and
𝑥
1
+ 𝑥
2
+ ⋯ + 𝑥
𝑛
= 𝑎.
Solution.
Let
max{𝑥
1
, 𝑥
2
, … , 𝑥
𝑛
} = 𝑥
𝑘
Then
Volume 04 Issue 06-2024
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VOLUME
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Pages:
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(2023:
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∑
𝑥
𝑖
𝑛−1
𝑖=1
𝑥
𝑖+1
= ∑
𝑥
𝑖
𝑘−1
𝑖=1
𝑥
𝑖+1
+ ∑
𝑥
𝑖
𝑛−1
𝑖=𝑘
𝑥
𝑖+1
≤ 𝑥
𝑘
∑
𝑥
𝑖
𝑘−1
𝑖=1
+𝑥
𝑘
.
∑
𝑥
𝑖
𝑘−1
𝑖=1
= 𝑥
𝑘
(𝑎 − 𝑥
𝑘
) ≤ ((
𝑥
𝑘
+ 𝑎 − 𝑥
𝑘
2
)
2
) =
𝑎
2
4
.
Equality is achieved, for example, when
𝑥
1
= 𝑥
2
=
𝑎
2
, 𝑥
3
= ⋯ = 𝑥
𝑛
= 0
Therefore, the largest value is
𝑎
2
4
.
C
ONCLUSION
As a rule, in practical problems it is necessary to
determine the largest and smallest values of a
function in a certain area. If the area is closed
and limited, then the differentiable function in
this area reaches is largest and smallest values
either at a stationary point or at the boundary
point of the area.
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