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Topic 8: Optimisation of functions of

several variables

Topic 8: Topic 8: OptimisationOptimisation of functions of of functions of

several variablesseveral variablesUnconstrained Unconstrained OptimisationOptimisation

((MaximisationMaximisation and and MinimisationMinimisation))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

AAt A, dY = 0And d2Y <0

Z

X

Y

0

B

At B, dY = 0And 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

fffdZffdXfYd

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 =−−= zxf x and 2. 02216 =−−= xzf z Solve simultaneously

2042 −=− xz

1622 −=− xz

So 162204 −=− xx 2=x

Subbing in x = 2 to eq. 1 (or 2): 122 −=− z and

6=z

There is 1 stationary point at (2,6)

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Second Order Conditions: 04 <−=xxf

02 <−=zzf

2−=xzf

Thus, fXX fZZ – (fXZ)2 = (-4.-2) – (-2) 2 = +8 –4 = +4 >0 Sufficient condition for Max (d2Y < 0). fxx < 0 and fXX fZZ – (fXZ)2 >0 So, Max at (2,6)

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Example 2(ii) xzzzxxy 85945100 22 ++−+−= Necessary Condition for max or min: 1. 0885 =++−= zxf x and 2. 08109 =++−= xzfz Solve simultaneously

zx 858 −= zx 1098 −=

zz 10985 −=− 42 =z

2=z Subbing in z = 2 to eq 1 (or 2) ( ) 112858 −=−=x 811−=x There is 1 stationary point at ( )2,811−

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Second Order Conditions 08 >=xxf

010 >=zzf 8=xzf

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) zzxxy 52

220 2

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−+−= Necessary Condition for max or min: 1. 020 =−= xf x and 2. 054 =−= zf z Thus, 20=x and 45=z There is 1 stationary point at ( )45,20 Second Order Conditions

01 <−=xxf 04 >=zzf

0=xzf 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 functions of several variables

OptimisationOptimisation of functions of functions of several variablesof several variablesEconomic 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 – 2QAand 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….

Π = TR - TC = PAQA + PBQB – TC

PAQA = (100 – 2QA)QA

PBQB = (100 – QB) QB

Π = 100QA – 2QA2 + 100QB – QB

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– 50QA – 50QB – ½ (QA+QB)2

= 50QA – 2QA2 + 50QB – QB

2 – ½ (QA+QB)2

Select QA and QB to max Π:

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if Π = 50QA – 2QA2 + 50QB – QB

2 – ½ (QA+QB)2

F.O.C. d Π =0ΠQA =50 - 4QA – ½ *2 (QA+QB) = 0

= 50 - 5QA – QB = 0 (1)ΠQB = 50 - 2QB – ½ *2 (QA+QB) = 0

= 50 - 3QB – QA = 0 (2)50 - 5QA – QB = 50 - 3QB – QA⇒ 2QA = QBThus, output at stationary point is

(QA,QB) = (71/7, 14 2/7 )

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Check Sufficient conditions for Max: d2Π <0

ΠQA = 50 - 5QA – QBΠQB = 50 - 3QB – QA

Then ΠQAQA = – 5 < 0

ΠQAQA. ΠQBQB – (ΠQAQB)2 >0

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

A and 14 2/7 units to country B.

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Example 2Profits and production

Max π = 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 π

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Necessary Condition: dπ = 0

πL = PQL – w = 0 , MPL = QL = w/PπK = PQK – r = 0 , MPK = QK = r/P

Sufficient Condition for a max, d2π <0

So πLL < 0 AND (πLL.πKK - πLK.πKL) > 0

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Max π = 2 K1/3L1/2 – L – 1/3 K{L*, K*}

Necessary condition for Max: dπ =0 (1) πL = K1/3L-1/2 – 1 = 0(2) πK = 2/3 K-2/3 L1/2 – 1/3 = 0Stationary 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 π?

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πL = K1/3L-1/2 – 1 πK = 2/3 K-2/3 L1/2 – 1/3

πLL = -1/2K1/3L-3/2 < 0 for all K and L

πKK = – 4/9 K–5/3L½

πKL = πLK = 1/3K–2/3L-½

For sufficient condition for a max,

Check d2π <0; πLL < 0 & (πLL.πKK - πLK.πKL)>0

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πLL.πKK =(-1/2K1/3L-3/2 ).( – 4/9 K–5/3L½ )= 4/18 . K–4/3L-1

πKL2 = (1/3K–2/3L-½). (1/3K–2/3L-½)

= 1/9K–4/3L-1

Thus, πLL.πKK > πKL.πLK since 4/18 > 1/9

So, (πLL.πKK - πKL.πLK) >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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