Optimisation of functions of several variables Unconstrained Optimisation, Slides of Econometrics and Mathematical Economics

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1

Optimisation of functions of

several variables

Unconstrained Optimisation 2

Recall……

Max Min X Y

3 First-Order / Necessary Condition:

dY/dX = f ¢ (X) = 0

Second-Order / Sufficent Condition:

d

2

Y/dX

2

= f ¢¢(X) if > 0 (Min)

if < 0 (Max)

Max Y = f (X) X* 4

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: d^2 Y = f ¢¢ (X).dX^2 >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

7 d 2 Y= fXX.dX 2

• fZZ.dZ 2
• 2fXZ dX.dZ Complete the Square (^22) 2 (^2) dZ f dZ f f^ f f dY f dX^ f XX XX ZZ XZ XX XX XZ ú ú û ù ê ê ë é (^) - ú^ + û ù ê ë =^ é^ + Sign Positive Definite To ensure d^2 Y > 0 and Min: fXX > 0 and fXXfZZ - fXZfZX > 0 Sign Negative Definite To ensure d^2 Y < 0 and Max: fXX < 0 and fXXfZZ - fXZfZX > 0 note: fXZ.fZX = (fXZ)^2 8 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 ….. d^2 Y > 0 fXX > d^2 Y > 0 fXX > 0 AND fXX fZZ – (fXZ)^2 > Sufficient For Max So required that ….. d^2 Y < 0 fXX < 0 d^2 Y < 0 fXX < 0 AND fXX fZZ – (fXZ) 2

fXX fZZ – (fXZ) 2 < Saddle Point

9 Examples Find all the maximum and minimum values of the functions: (i) y^^10 20 x^2 x^16 z z^2 xz = + -^2 + -^2 - Necessary Condition for max or min:

1. fx =^20 -^4 x -^2 z =^0 and
2. fz =^16 -^2 z -^2 x =^0 Solve simultaneously
   • 2 z = 4 x - 20
   • 2 z = 2 x - 16 So 4 x -^20 =^2 x -^16 x = 2 Subbing in x = 2 to eq. 1 (or 2):
   • 2 z =- (^12) and z = 6 There is 1 stationary point at (2,6) 10 Second Order Conditions: fxx =- 4 < 0 fzz =- 2 < 0 fxz =- 2 Thus, fXX fZZ – (fXZ) 2 = (-4.-2) – (-2) 2 = +8 – 4 = +4 > Sufficient condition for Max ( d 2 Y < 0). fxx < 0 and fXX fZZ – (fXZ) 2

So, Max at (2,6)

13

Example 3

(iii) z z y x^ x 2 5 2 20 2 2 = - + - Necessary Condition for max or min:

1. f^ x =^20 - x =^0 and
2. fz =^4 z -^5 =^0 Thus, x^ =^20 and z =^54 There is 1 stationary point at (^20 ,^54 ) Second Order Conditions fxx =- 1 < 0 fzz = 4 > 0 fxz = 0 fXX fZZ – (fXZ)^2 = (-1.4) – (0^2 ) = - 4 < 0 Since, fxx < 0 and fXX fZZ – (fXZ)^2 <0, Saddle Point at (^20 ,^54 ) 14

Optimisation of functions of several variables

Economic Applications

15

Example 1

A 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? 16

Objective Function to Max is Profit….

P = TR - TC = PAQA + PBQB – TC PAQA = ( 100 – 2 QA)QA PBQB = ( 100 – QB) QB P = 100 QA – 2 QA^2 + 100 QB – QB^2

• 50 QA – 50 QB – ½ (QA+QB)^2 = 50 QA – 2 QA^2 + 50 QB – QB^2 – ½ (QA+QB)^2 Select QA and QB to max P :

19

Example 2

Profits and production Max p = 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 p 20 Necessary Condition: dp = 0 pL = PQL – w = 0 , MPL = QL = w/P pK = PQK – r = 0 , MPK = QK = r/P Sufficient Condition for a max, d^2 p < So pLL < 0 AND (pLL.pKK - pLK.pKL) > 0

21 Max p = 2 K1/3L1/2^ – L – 1/3 K {L, K}

Necessary condition for Max: d p =

(1) pL = K1/3L-1/2^ – 1 = 0 (2) pK = 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) / 3 K– 2/3K1/3 (^) = 1 / 3. Re-arranging K– 1/3 (^) = ½ and so K 1/3 (^) = 2 = L½. Thus, K* =2^3 = 8. And so L* = 2^2 = 4. NOW, let Q = K1/3L1/2, P = 2, w = 1, r =1/ Find the values of L & K that max p? 22 pL = K1/3L-1/2^ – 1 pK = 2 / 3 K-2/3^ L1/2^ – 1 / 3 pLL = -^1 / 2 K1/3L-3/2^ < 0 for all K and L pKK = – 4 / 9 K–5/3L½ pKL = pLK = 1 / 3 K–2/3L-½ For sufficient condition for a max, Check d^2 p <0; p LL < 0 & ( p LL. p KK - p LK. p KL)>