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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

222

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

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Second Order Conditions 08 xxf

010 zzf

8xzf fXX fZZ – (fXZ)2 =

(8.10) – (82) = 80 – 64 = 16 ….>0 Sufficient condition for Min(d2Y > 0). fxx > 0 and fXX fZZ – (fXZ)2 >0

So, Min at 2,811

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Example 3(iii) zz

xxy 52

220 2

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Necessary Condition for max or min:

1. 020 xf x and

2. 054 zf z Thus, 20x and 45z There is 1 stationary point at 45,20 Second Order Conditions

01xxf 04 zzf

0xzf fXX fZZ – (fXZ)2 = (-1.4) – (02) = - 4 < 0 Since, fxx < 0 and fXX fZZ – (fXZ)2 <0, Saddle Point at 45,20

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Optimisation of Optimisation of functions of several functions of several

variablesvariables

Optimisation of Optimisation of functions of several functions of several

variablesvariables

Economic ApplicationsEconomic Applications

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Example 1A firm can sell its product in two

countries, A and B, where demand in country A is given by PA = 100 – 2QA and in country B is PB = 100 – QB.

It’s total output is QA + QB, which it can produce at a cost of

TC = 50(QA+QB) + ½ (QA+QB)2

How much will it sell in the two countries assuming it maximises profits?

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Objective Function to Max is Profit….

= TR - TC = PAQA + PBQB – TC

PAQA = (100 – 2QA)QA

PBQB = (100 – QB)

QB

  = 100QA – 2QA2 + 100QB – QB

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– 50QA – 50QB – ½ (QA+QB)

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= 50QA – 2QA2 + 50QB – QB

2 – ½

(QA+QB)2

Select QA and QB to max :

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if = 50QA – 2QA2 + 50QB – QB

2 – ½

(QA+QB)2

F.O.C. d =0

QA =50 - 4QA – ½ *2 (QA+QB) = 0

= 50 - 5QA – QB = 0 (1)

 QB = 50 - 2QB – ½ *2 (QA+QB) = 0

= 50 - 3QB – QA = 0 (2)

50 - 5QA – QB = 50 - 3QB – QA

2QA = QB

Thus, output at stationary point is (QA,QB) = (71/7, 14 2/7 )

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Check Sufficient conditions for Max: d2 <0

QA = 50 - 5QA – QB

QB = 50 - 3QB – QA

Then QAQA = – 5 < 0

QAQA. QBQB – (QAQB)2 >0

(–5 * –3)) – (-1) 2 = 14 > 0 Max So firm max profits by selling 71/7 units to

country A and 14 2/7 units to country B.

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Example 2 Profits and production

Max = PQ(L, K) – wL - rK

{L*, K*}

Total Revenue = PQ

Expenditure on labour L = wL

Expenditure on Capital K = rK

Find the values of L & K that max

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Necessary Condition: d = 0

L = PQL – w = 0 , MPL = QL = w/P

K = PQK – r = 0 , MPK = QK = r/P

 Sufficient Condition for a max, d2 <0

So LL < 0 AND (LL.KK - LK.KL) > 0

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Max = 2 K1/3L1/2 – L – 1/3 K{L*, K*}

Necessary condition for Max: d =0

(1) L = K1/3L-1/2 – 1 = 0

(2) K = 2/3 K-2/3 L1/2 – 1/3 = 0

Stationary point at [L*, K*] = [4, 8] note: to solve, from eq1: L½ = K1/3 . Substituting into eq2 then, 2/3K

– 2/3K1/3 = 1/3. Re-arranging K– 1/3 = ½ and so K 1/3 = 2 = L½.

Thus, K* =23= 8. And so L* = 22 = 4.

NOW, let Q = K1/3L1/2, P = 2, w = 1, r =1/3

Find the values of L & K that max ?

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L = K1/3L-1/2 – 1

K = 2/3 K-2/3 L1/2 – 1/3

LL = -1/2K

1/3L-3/2 < 0 for all K and L

 KK = – 4/9 K–5/3L½

KL = LK = 1/3K–2/3L-½

For sufficient condition for a max,

Check d2 <0; LL < 0 & (LL.KK - LK.KL)>0

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LL.KK =(-1/2K1/3L-3/2 ).( – 4/9 K

–5/3L½ )

= 4/18 . K–4/3L-1

KL2 = (1/3K

–2/3L-½). (1/3K–2/3L-½)

= 1/9K–4/3L-1

 Thus, LL.KK >

KL.LK since 4/18 > 1/9

 So, (LL.KK - KL.LK) >0 for all values of K

& L Profit max at stationary point [L*, K*] = [4, 8]

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Unconstrained Optimisation –

Functions of Several Variables

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