Topic 8: Optimisation of functions of several variables

1. Topic 8: Optimisation of functions of several variables Unconstrained Optimisation (Maximisation and Minimisation) Jacques (4th Edition): 5.4

2. Recall…… Y Max Min X

3. Max Y = f (X) X*

4. Re-writing in terms of total differentials….

6. Max Y = f (X, Z)[X*, Z*] Necessary Condition: dY = fX.dX + fZ.dZ = 0 so 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

8. Optimisation - A summing Up…

9. Examples

11. Example 2

13. Example 3

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….  = TR - TC = PAQA + PBQB – TC PAQA = (100 – 2QA)QA PBQB=(100 – QB)QB  = 100QA – 2QA2 + 100QB – QB2 – 50QA – 50QB – ½ (QA+QB)2 = 50QA – 2QA2 + 50QB – QB2– ½ (QA+QB)2 Select QA and QB to max :

17. if  = 50QA – 2QA2 + 50QB – QB2– 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 )

18. 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.

19. 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 

20. 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

21. NOW, let Q = K1/3L1/2, P = 2, w = 1, r =1/3 Find the values of L & K that max ? 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.

22. For sufficient condition for a max, Check d2 <0; LL < 0 & (LL.KK - LK.KL)>0 L = K1/3L-1/2 – 1 K = 2/3 K-2/3 L1/2 – 1/3 LL = -1/2K1/3L-3/2 < 0 for all K and L KK = – 4/9 K–5/3L½ KL = LK = 1/3K–2/3L-½

23. 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]

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