拉格蘭吉乘數

第三十一單元

Constrained Optimization ProblemA One-variable problem

Lagrange

Joseph-Louis Lagrange

(1736-1813)

Mathematician who developed many fundamental techniques in the calculus of variations, including the method that bears his name.

Geometric Interpretation

Lagrange Method

Lagrange Theorem

0 0

Suppose ( , , ) and ( , , ) are functions

with continuous first partial derivatives and

( , , ) 0 on the surface ( , , ) 0.

Suppose the maximum(minimum) value of

( , , ) subject to the constraint ( , , ) 0

occurs at ( , ,

f x y z g x y z

g x y z g x y z

f x y z g x y z

x y

Ñ ¹ =

=

0

0 0 0 0 0 0

), then

( , , ) ( , , )

for some constant

z

f x y z g x y zl

l

Ñ = Ñ

Procedure for the Method of Lagrange

Suppose and satify the hypothesis of Lagrange

theorem , ( , ) has extremum subject to the

constraint ( , ) . Then to find the extreme

values process is :

f g

f x y

g x y c=

(1) , ,y yx xf g f colve gS gll = ==

(2) Evaluate at all points found in (1),

the extremum must among these values

f

ExampleFind the point to the plane 2 1x y z+ + =that is nearest to the origin .

( , , )x y z2 2 2x y z+ +

ExampleFind the point to the plane 2 1x y z+ + =that is nearest to the origin .

Solution :2 2 2Find the smallest value of ( , , )f x y z x y z= + +

subjct to the constraint 2 1x y z+ + =2 , 2 , 2x y zf x f y f z= = =(1)

2 , 1 , 1x y zg g g= = =

2 2

2 1

2 1

x

y

z

l

l

l

ì = ×ïïïï = ×íïï = ×ïïî

2 2x y zl = = =

Example(continued)2 2x y zl = = =

2 1x y z+ + = ÞQ1

2 1 3 1,2 2 3

l ll l l+ + = Þ = =

1 1 1, ,

3 6 6x y z= = =

2 2 21 1 1 1 1 1 6 1

, ,3 6 6 3 6 6 36 6

fæ ö æö æö æö÷ ÷ ÷ ÷ç ç ç ç= + + = =÷ ÷ ÷ ÷ç ç ç ç÷ ÷ ÷ ÷÷ ÷ ÷ ÷ç ç ç çè ø è ø è ø è ø

1The shortest distance is

6

Example(continue)

(2) By Cauchy-Schwartz Inequality

2(2 )x y z+ + 2 2 2( )x y z£ + + 2 2 2(2 1 1 )+ +

2 2 2 21 ( )x y z£ + + 6×

2 2 2 1( )

6x y z+ + ³

2 2 2 1

6x y z+ + ³

Example

( , ) 2yf x y y=-

2 2Find the largest values of ( , ) 1

subject to the constraint 1 with 0, 0

f x y x y

x y x y

= - -

+ = ³ ³

Solution :the constraint is 1, let ( , ) 1x y g x y x y+ = = + -Q

( , ) 2xf x y x=-

( , ) 1xg x y = ( , ) 1yg x y =

2 1

2 1

1

x

y

x y

l

l

ì - = ×ïïïï - = ×íïï + =ïïî

Example(continued)

(2)

(1

(3)

1

1

1

)2

2

x

y

x y

l

l

ì - = ×ïïïï - = ×íïï + =ïïî

L

L L

L

L L(2)1),(By x yÞ =

(3),1 1

Substituting into we get , , 12 2

x y x y l= = = =-2 2

1 1 1 1 1, 1

2 2 2 2 2fæ ö æö æö÷ ÷ ÷ç ç ç= - - =÷ ÷ ÷ç ç ç÷ ÷ ÷ç ç çè ø è ø è ø

2 2(1,0) 1 1 0 0f = - - =2 2(0,1) 1 0 1 0f = - - =

1 1 1Maximum value ,

2 2 2atæ ö÷ç\ = ÷ç ÷çè ø

( ) ( )Minimum value 0 1,0 and 0,1at\ =

Animate plot 3d

ExampleMaximize ( , ) subject to 20f x y xy x y= + =

Solution(1) :,x yf y f x= =1

1

20

y

x

x y

l

l

ì = ×ïïïï = ×íïï + =ïïî2 20 , 10l lÞ = =

Solution(2) :

2

x yxy

+³

20

2xy³

10 xy³

100xy £10 , 10x y= =

100xyÞ =

Example (Optimization inside a Region)2 2Find the extreme of ( , ) 2 2 1f x y x y x= + - +

2 2subject to the constraint 10x y+ £

Solution:Consider the points on the c(1) ircle

2 2( , ) 2 2 1f x y x y x= + - +2 2( , ) 10 0g x y x y= + - =

ìïïïïíïïïïî

2 2x - = l (2 )x (1)L L

4y= l (2 )y (2)L L2 2 10 0x y+ - = (3)L L

By (2) 2l = 2 2x - = 2 (2 )x 1xÞ =-

3yÞ =±

Solution(continued) Consider the poin insidets the ci(2) rcle

2 2xf x= -ìïïïïíïïïïî2yf y=

0=

0=(1,0) critical pointÞ

xx xy

xy yy

f fD

f f= =

2 0

0 24 0= >

0xxf > (1,0) relative minimumÞCombin(3) (1e ), (2)

( 1, 3) 22f - ± =(1,0) 0f =

maximumL L

minimumL L

單元結語

有限制條件的極值問題除了特定的狀況可使用歌西不等式、算 幾的方法外，拉格蘭吉乘子法都可以算。

拉格蘭吉乘子法不限於於只有一個限制條件，兩個、三個限制條件都可以解，在此為避免計算過於冗長，所舉得例題只做一個限制條件的問題。

³

1

2

3

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