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

2 Recall…… Max Min X Y

3 Max Y = f (X) X*

4 Re-writing in terms of total differentials….

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

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8 Optimisation - A summing Up …

9 Examples

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11 Example 2

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13 Example 3

14 Optimisation of functions of several variables Economic Applications 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 P A = 100 – 2Q A and in country B is P B = 100 – Q B. It’s total output is Q A + Q B, which it can produce at a cost of TC = 50(Q A +Q B ) + ½ (Q A +Q B ) 2 How much will it sell in the two countries assuming it maximises profits?

16 Objective Function to Max is Profit….  = TR - TC = P A Q A + P B Q B – TC P A Q A = ( 100 – 2Q A )Q A P B Q B = ( 100 – Q B ) Q B  = 100Q A – 2Q A Q B – Q B 2 – 50Q A – 50Q B – ½ (Q A +Q B ) 2 = 50Q A – 2Q A Q B – Q B 2 – ½ (Q A +Q B ) 2 Select Q A and Q B to max  :

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

18 Check Sufficient conditions for Max: d 2  <0  QA = Q A – Q B  QB = Q B – Q A Then  QAQA = – 5 < 0  QAQA.  QBQB – (  QAQB ) 2 >0 (–5 * –3)) – (-1) 2 = 14 > 0 Max So firm max profits by selling 7 1 / 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 = PQ L – w = 0, MPL = Q L = w/P  K = PQ K – r = 0, MPK = Q K = r/P Sufficient Condition for a max, d 2  <0 So  LL 0

21 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 eq1: L ½ = K 1/3. Substituting into eq2 then, 2 / 3 K – 2/3 K 1/3 = 1 / 3. Re-arranging K – 1/3 = ½ and so K 1/3 = 2 = L ½. Thus, K* =2 3 = 8. And so L* = 2 2 = 4. NOW, let Q = K 1/3 L 1/2, P = 2, w = 1, r =1/3 Find the values of L & K that max  ?

22  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 L ½  KL =  LK = 1 / 3 K –2/3 L -½ For sufficient condition for a max, Check d 2  0

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

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