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

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2 Recall…… Max Min X Y

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3 Max Y = f (X) X*

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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 …

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9 Examples

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11 Example 2

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13 Example 3

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14 Optimisation of functions of several variables Economic Applications Economic Applications

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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?

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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 2 + 100Q B – Q B 2 – 50Q A – 50Q B – ½ (Q A +Q B ) 2 = 50Q A – 2Q A 2 + 50Q B – Q B 2 – ½ (Q A +Q B ) 2 Select Q A and Q B to max :

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17 if = 50Q A – 2Q A 2 + 50Q 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 = 50 - 5Q A – Q B = 0(1) QB = 50 - 2Q B – ½ *2 (Q A +Q B ) = 0 = 50 - 3Q B – Q A = 0(2) 50 - 5Q A – Q B = 50 - 3Q B – Q A 2Q A = Q B Thus, output at stationary point is (Q A, Q B ) = (7 1 / 7, 14 2 / 7 )

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

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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

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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

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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 ?

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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

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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]

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

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