Topic 8: Optimisation of functions of several variables

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Topic 8 Optimisation of functions of several
variables

• Unconstrained Optimisation
• (Maximisation and Minimisation)
• Jacques (4th Edition) 5.4

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Recall
Y
Max
Min
X
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Max Y f (X) X
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Re-writing in terms of total differentials.
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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 ?
• gt0 for Min
• lt0 for Max
• Sign Positive Definite ? Min
• Sign Negative Definite ? Max

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Optimisation - A summing Up
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Examples
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Example 2
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Example 3
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Optimisation of functions of several variables

• Economic Applications

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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.
• Its total output is QA QB, which it can
  produce at a cost of
• TC 50(QAQB) ½ (QAQB)2
• How much will it sell in the two countries
  assuming it maximises profits?

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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 (QAQB)2 50QA 2QA2 50QB
QB2 ½ (QAQB)2 Select QA and QB to max ?
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if ? 50QA 2QA2 50QB QB2 ½
(QAQB)2

• F.O.C. d ? 0
• ?QA 50 - 4QA ½ 2 (QAQB) 0
• 50 - 5QA QB 0 (1)
•  ?QB 50 - 2QB ½ 2 (QAQB) 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 )

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Check Sufficient conditions for Max d2? lt0

• ?QA 50 - 5QA QB
• ?QB 50 - 3QB QA
• Then
• ?QAQA 5 lt 0
• ?QAQA. ?QBQB (?QAQB)2 gt0
• (5 3)) (-1) 2 14 gt 0 Max
•  So firm max profits by selling 71/7 units to
  country A and 14 2/7 units to country B.

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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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• Necessary Condition d? 0
• ?L PQL w 0 , MPL QL w/P
• ?K PQK r 0 , MPK QK r/P
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• Sufficient Condition for a max, d2? lt0
• So ?LL lt 0 AND (?LL.?KK - ?LK.?KL) gt 0

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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.

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For sufficient condition for a max, Check d2?
lt0 ?LL lt 0 (?LL.?KK - ?LK.?KL)gt0

• ?L K1/3L-1/2 1
• ?K 2/3 K-2/3 L1/2 1/3
• ?LL -1/2K1/3L-3/2 lt 0 for all K and L
•  ?KK 4/9 K5/3L½
• ?KL ?LK 1/3K2/3L-½

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• ?LL.?KK (-1/2K1/3L-3/2 ).( 4/9 K5/3L½ )
• 4/18 . K4/3L-1
• ?KL2 (1/3K2/3L-½). (1/3K2/3L-½)
• 1/9K4/3L-1
•  Thus, ?LL.?KK gt ?KL.?LK since 4/18 gt 1/9
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• So, (?LL.?KK - ?KL.?LK) gt0 for all values of K
  L
•  Profit max at stationary point
• L, K 4, 8

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Unconstrained Optimisation Functions of
Several Variables

• Self-Assessment Questions on Website
• Tutorial problem sheets
• Pass Exam Papers
• Examples in the Textbook