1 Topic 8: Topic 8: Optimisation of Optimisation of functions of several functions of several variables variables Unconstrained Optimisation Unconstrained Optimisation (Maximisation and Minimisation) (Maximisation and Minimisation) Jacques (4th Edition): 5.4 Jacques (4th Edition): 5.4
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1 Topic 8: Optimisation of functions of several variables Unconstrained Optimisation (Maximisation and Minimisation) Jacques (4th Edition): 5.4.
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Topic 8: Topic 8: Optimisation of functions Optimisation of functions
of several variablesof several variables
Topic 8: Topic 8: Optimisation of functions Optimisation of functions
of several variablesof several variablesUnconstrained OptimisationUnconstrained Optimisation
(Maximisation and Minimisation)(Maximisation and Minimisation)
Jacques (4th Edition): 5.4Jacques (4th Edition): 5.4
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Recall……Max
Min
X
Y
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First-Order / Necessary Condition:
dY/dX = f (X) = 0 Second-Order / Sufficent Condition:
d2Y/dX2 = f (X) if > 0 (Min) if < 0 (Max)
Max Y = f (X) X*
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Re-writing in terms of total differentials….
Max Y = f (X) X* Necessary Condition:
dY = f (X).dX = 0 , so it must be that f (X)= 0 Sufficent Condition:
d2Y = f (X).dX2 >0 for Min <0 for Max For Positive Definite, (Min ), it must be that f > 0 Negative Definite, (Max), it must be that f < 0
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Z
Y
X 0
A At A, dY = 0 And d2Y <0
Z
X
Y
0
B
At B, dY = 0 And d2Y >0
Max Y = f(X,Z) [X*Z*]
Min Y = f(X,Z) [X*Z*]
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Max Y = f (X, Z)[X*, Z*]
Necessary Condition:dY = fX.dX + fZ.dZ = 0so it must be that fX = 0 AND fZ = 0
Sufficient Condition:
d2Y= fXX.dX2 +fZX dZ.dX + fZZ.dZ2 + fXZ .dXdZ
….and since fZX = fXZ
d2Y= fXX.dX2 + fZZ.dZ2 + 2fXZ dX.dZ ?
>0 for Min<0 for Max
Sign Positive Definite Min Sign Negative Definite Max
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d2Y= fXX.dX2 + fZZ.dZ2 + 2fXZ dX.dZ
Complete the Square
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2 dZf
fffdZ
f
fdXfYd
XX
XZZZXX
XX
XZXX
Sign Positive Definite To ensure d2Y > 0 and Min:
00 ZXXZZZXXXX ffffandf Sign Negative Definite To ensure d2Y < 0 and Max:
00 ZXXZZZXXXX ffffandf note: fXZ.fZX = (fXZ)2
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Optimisation - A summing Up… Condition Y = f(X) Y = f(X,Z)
Neccesary
So required that…….
dY = 0
fX = 0
dY = 0
fX = 0 AND fZ = 0
Sufficient For Min
So required that …..
d2Y > 0
fXX >0
d2Y > 0
fXX > 0 AND fXX fZZ – (fXZ)2 >0
Sufficient For Max
So required that …..
d2Y < 0
fXX < 0
d2Y < 0
fXX < 0 AND fXX fZZ – (fXZ)2 >0
fXX fZZ – (fXZ)2 <0 Saddle Point
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ExamplesFind all the maximum and minimum values of the functions:
(i) xzzzxxy 21622010 22
Necessary Condition for max or min: 1. 02420 zxfx and
2. 02216 xzf z Solve simultaneously
2042 xz
1622 xz
So 162204 xx 2x
Subbing in x = 2 to eq. 1 (or 2): 122 z and
6z
There is 1 stationary point at (2,6)
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S e c o n d O r d e r C o n d i t i o n s : 04 xxf
02 zzf
2xzf
T h u s , f X X f Z Z – ( f X Z ) 2 = ( - 4 . - 2 ) – ( - 2 ) 2 = + 8 – 4 = + 4 > 0 S u f f i c i e n t c o n d i t i o n f o r M a x ( d 2 Y < 0 ) . f x x < 0 a n d f X X f Z Z – ( f X Z ) 2 > 0 S o , M a x a t ( 2 , 6 )
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Example 2( i i ) xzzzxxy 85945100 22 N e c e s s a r y C o n d i t i o n f o r m a x o r m i n :
1 . 0885 zxf x a n d 2 . 08109 xzf z S o l v e s i m u l t a n e o u s l y
zx 858 zx 1098
zz 10985 42 z
2z S u b b i n g i n z = 2 t o e q 1 ( o r 2 )
112858 x 811x T h e r e i s 1 s t a t i o n a r y p o i n t a t 2,811