Topic 8 Optimisation of functions of several variables

Topic 8: Optimisation of functions of several variables Unconstrained Optimisation (Maximisation and Minimisation) Jacques (4 th Edition): 5. 4 1

Recall…… Y Max Min X 2

Max Y = f (X) X* 3

Re-writing in terms of total differentials…. 4

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Max Y = f (X, Z) [X*, Z*] Necessary Condition: d. Y = f. X. d. X + f. Z. d. Z = 0 so it must be that f. X = 0 AND f. Z = 0 Sufficient Condition: d 2 Y= f. XX. d. X 2 +f. ZX d. Z. d. X + f. ZZ. d. Z 2 + f. XZ. d. Xd. Z …. and since f. ZX = f. XZ d 2 Y= f. XX. d. X 2 + f. ZZ. d. Z 2 + 2 f. XZ d. X. d. Z ? >0 for Min <0 for Max Sign Positive Definite Min Sign Negative Definite Max 6

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Optimisation - A summing Up… 8

Examples 9

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Example 2 11

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Example 3 13

Optimisation of functions of several variables Economic Applications 14

Example 1 A firm can sell its product in two countries, A and B, where demand in country A is given by PA = 100 – 2 QA 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? 15

Objective Function to Max is Profit…. = TR - TC = PAQA + PBQB – TC PAQA = (100 – 2 QA)QA PBQB = (100 – QB) QB = 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 : 16

if = 50 QA – 2 QA 2 + 50 QB – QB 2 – ½ (QA+QB)2 F. O. C. d =0 QA =50 - 4 QA – ½ *2 (QA+QB) = 0 = 50 - 5 QA – QB = 0 (1) QB = 50 - 2 QB – ½ *2 (QA+QB) = 0 = 50 - 3 QB – QA = 0 (2) 50 - 5 QA – QB = 50 - 3 QB – QA 2 QA = QB Thus, output at stationary point is (QA, QB) = (71/7, 14 2/7 ) 17

Check Sufficient conditions for Max: d 2 <0 QA = 50 - 5 QA – QB QB = 50 - 3 QB – 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. 18

Example 2 Profits and production Max = PQ(L, K) – w. L - r. K {L*, K*} Total Revenue = PQ Expenditure on labour L = w. L Expenditure on Capital K = r. K Find the values of L & K that max 19

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, d 2 <0 So LL < 0 AND ( LL. KK - LK. KL) > 0 20

NOW, let Q = K 1/3 L 1/2, P = 2, w = 1, r =1/3 Find the values of L & K that max ? Max = 2 K 1/3 L 1/2 – L – 1/3 K {L*, K*} Necessary condition for Max: d =0 (1) L = K 1/3 L-1/2 – 1 = 0 (2) K = 2/3 K-2/3 L 1/2 – 1/3 = 0 Stationary point at [L*, K*] = [4, 8] note: to solve, from eq 1: L = K. Substituting into eq 2 then, ½ 2/ 3 K 1/3 – 2/3 K 1/3 = 1/3. Re-arranging K– 1/3 = ½ and so K 1/3 = 2 = L½. Thus, K* =23= 8. And so L* = 22 = 4. 21

For sufficient condition for a max, Check d 2 <0; LL < 0 & ( LL. KK - LK. KL)>0 L = K 1/3 L-1/2 – 1 K = 2/3 K-2/3 L 1/2 – 1/3 LL = -1/2 K 1/3 L-3/2 < 0 for all K and L KK = – 4/9 K– 5/3 L½ KL = LK = 1/3 K– 2/3 L-½ 22

LL. KK =(-1/2 K 1/3 L-3/2 ). ( – 4/9 K– 5/3 L½ ) = 4/18. K– 4/3 L-1 KL 2 = (1/3 K– 2/3 L-½) = 1/9 K– 4/3 L-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] 23

Unconstrained Optimisation – Functions of Several Variables • Self-Assessment Questions on Website • Tutorial problem sheets • Pass Exam Papers • Examples in the Textbook 24