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EECM 3714

Lecture 9: Unit 9

Constrained Optimisation
Renshaw, Ch
...

• Suppose that we want to maximise/minimise 𝑓(𝑥, 𝑦) subject to a constraint, 𝑔(𝑥, 𝑦)

• max 𝑓(𝑥, 𝑦), s
...
𝑔(𝑥, 𝑦)
• min 𝑓(𝑥, 𝑦), s
...
𝑔(𝑥, 𝑦)

• Where: 𝑓(𝑥, 𝑦) is the objective function (the function to be minimised or maximised); and 𝑔(𝑥, 𝑦)
is the constraint
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2, Fig 16
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5

THE LAGRANGE MULTIPLIER (LM) METHOD
• Suppose that you have to optimize (min/max) 𝑓(𝑥, 𝑦), s
...
𝑐 = 𝑔(𝑥, 𝑦)

• Rewrite the constraint, so that 𝑐 − 𝑔 𝑥, 𝑦 = 0
• Then, define a new function, the Lagrangian: 𝑉 = 𝑓(𝑥, 𝑦)+λ(𝑐 − 𝑔 𝑥, 𝑦 )
• Where f = the objective function
• λ = the Lagrange Multiplier (LM)
• 𝑐 − 𝑔 𝑥, 𝑦 = the constraint

• Note that the Lagrangian can also be written as 𝑉 = 𝑓 𝑥, 𝑦 − λ(𝑔 𝑥, 𝑦 − 𝑐)
• To find the values of 𝑥, 𝑦 that max or min V,
𝜕𝑉 𝜕𝑉 𝜕𝑉

• Find the first partial derivatives 𝜕𝑥 ; 𝜕𝑦 ; 𝜕𝜆

• Then set these partial derivatives equal to zero,
𝜕𝑉

𝜕𝑉

• i
...
𝜕𝑥 = 𝜕𝑦 =

𝜕𝑉
𝜕𝜆

=0

• Solve for 𝑥, 𝑦, 𝜆
• This is the first-order condition for constrained optimisation
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2-16
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526-7)

ECONOMIC APPLICATIONS
• Cost minimisation subject to a production quota/constraint: 𝑚𝑖𝑛 𝐶 𝐾, 𝐿 s
...
𝑞𝑜 = 𝑔(𝐾, 𝐿)
• Dual problem: maximise output subject to a cost constraint: 𝑚𝑎𝑥 𝑞 𝐾; 𝐿 s
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𝐶0 = 𝐶(𝐾, 𝐿)
• Maximise utility subject to a budget constraint: 𝑚𝑎𝑥 𝑈 𝑥, 𝑦 s
...
𝑀0 = 𝑔(𝑥, 𝑦)

• Dual problem: minimize expenditure subject to a utility constraint: 𝑚𝑖𝑛 𝑀 𝑥; 𝑦 s
...
𝑈0 =
𝑈(𝑥, 𝑦)

COST MINIMISATION
• Suppose that a firm’s cost function is 𝐶 = 𝑤𝐿 + 𝑟𝐾, while its production function is 𝑞 = 𝑞(𝐾, 𝐿)

• Now suppose that the firm needs to meet a production quota, 𝑞0
• The firm’s production quota/constraint is then 𝑞0 = 𝑞 𝐾, 𝐿 ⟹ 𝑞0 − 𝑞 𝐾, 𝐿 = 0
• This can also be written as 𝑞 𝐾, 𝐿 − 𝑞0 = 0

• Now suppose that the firm wants to minimise its costs subject to the production quota
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9 & Fig 16
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e
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Use the Lagrange Multiplier method to find the values of 𝐿, 𝐾, 𝜆 that minimise the firm’s total

cost, subject to the constraint 100 = 5𝐾

1Τ 2Τ
3
3

𝐿

b
...


c
...
e
...
t
...


1Τ 2Τ
3
3

𝐿


...
(1)
………
...
(3)

• Divide equation (1) by equation (2) – LHS of (1) by LHS of (2) and
4
2

RHS of (1) by RHS of (2) → =
• → 2 = 2𝐾

1Τ −(−2Τ )
3
3

𝐿−

1Τ −2Τ
3
3

10Τ 𝜆𝐾 1ൗ3 𝐿 −1ൗ3
3
5Τ 𝜆𝐾 −2ൗ3 𝐿 2ൗ3
3

→ 2 = 2𝐾𝐿−1 ; ∴ 𝐾 = 𝐿

• Substitute K (or L) into equation (3)

• → 100 = 5𝐾

1Τ
3

𝐾

2Τ
3

⟹ 100 = 5𝐾, ∴ 𝐾 = 20 ⟹ 𝐿 = 20

• Solve for 𝜆: Plug 𝐾 = 𝐿 = 20 into eq
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(2)]: 4 =
10Τ 𝜆(20)1Τ3 (20)−1Τ3 , ∴ 𝜆 = 1
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2 (in this case, = MC

SOLUTION:
• Show that W/R = MRTS
𝑀𝑃𝐿
𝑀𝑃𝐾
1
1
𝑀𝑃𝐿 = 10Τ3 𝐾 Τ3 𝐿− Τ3
2
2
𝑀𝑃𝐾 = 5Τ3 𝐾 − Τ3 𝐿 Τ3

• 𝑀𝑅𝑇𝑆 =
•
•
•

𝑀𝑃𝐿
𝑀𝑃𝐾

=

10Τ 𝐾 1ൗ3 𝐿 −1ൗ3
3
5Τ 𝐾 −2ൗ3 𝐿 2ൗ3
3

=

• Since 𝐾 = 𝐿 = 20 →

2𝐾
𝐿

= 2𝐾𝐿−1

𝑀𝑃𝐿
𝑀𝑃𝐾

= 2 20 20−1 = 2

• Given 𝐶 = 4𝐿 + 2𝐾 → 𝑤=4 and 𝑟 = 2
•

𝑤
𝑟

• ∴

4
2

= =2
𝑤
𝑟

= 𝑀𝑅𝑇𝑆 = 2 if 𝐿 = 𝐾 = 20

OUTPUT MAXIMISATION
This is known as the dual problem for the cost minimisation problem - work in opposite direction
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3 𝐾 0
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Also suppose that 𝑤 = 3; 𝑟 =
15; 𝐶 = 150
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Find the values of 𝐾, 𝐿 that maximise production, subject to the cost constraint, i
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max
(𝐿0
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7 ) s
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3𝐿 + 15𝐾 = 150
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3 𝐾 0
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3 𝐾 0
...
3𝐿−0
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7 − 3𝜆 = 0 ⟹ 0
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7 𝐾 0
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7𝐿0
...
3 − 15𝜆 = 0 ⟹ 0
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3 𝐾 −0
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3𝐿−0
...
7
0
...
3 𝐾−0
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(1)
……
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(3)

3𝜆

= 15𝜆
7

= 5 ⟹ 15𝐾 = 7𝐿 → 𝐾 = 15 𝐿 (note that, here you can also make L the subject of the formula)

• Plug K into (3): 150 − 3𝐿 − 15

7
𝐿
15

=0

• ⟹ 150 − 3𝐿 − 7𝐿 = 0, ∴ 𝐿 = 15, 𝐾 = 7

• 0
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7 70
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059
• Thus, one unit increase in the cost constraint will lead to a 0
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The price of x is 𝑝𝑥 ,
while the price of y is 𝑝𝑦
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This means that the consumer’s
budget constraint is 𝑀0 = 𝑝𝑥 𝑥 + 𝑝𝑦 𝑦
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14 & Fig 16
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• Solve for 𝑥, 𝑦 and 𝜆

• In equilibrium,

𝑀𝑈𝑥
𝑝𝑥

=

𝑀𝑈𝑦

𝑝𝑦

⟹

𝑀𝑈𝑥
𝑀𝑈𝑦

=

𝑝𝑥
𝑝𝑦

⟹ 𝑀𝑅𝐶𝑆 =

𝑝𝑥
𝑝𝑦

• 𝜆 tells us by how many utils the utility function will change is the constraint increases by R1 (i
...

we have more income to spend on x and y)

EXAMPLE 1: UTILITY MAXIMISATION
• A consumer’s utility function is 𝑈 = 𝑥

1Τ
3

𝑦

2Τ
3


...

a
...
e
...


Show that

2Τ
3

s
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120 = 4𝑥 + 2𝑦
...
(1)

+ 𝜆 120 − 4𝑥 − 2𝑦 = 𝑥
𝜕𝑉
𝜕𝜆

2Τ
3

2Τ
3

=0
1

2

2

1Τ
3

− 4𝜆 = 0 ⟹ 3 𝑥 − Τ3 𝑦

1

𝑦 − Τ3 − 2𝜆 = 0 ⟹ 3 𝑥

1

𝑦 − Τ3 = 2𝜆………
...
(3)

4𝜆

= 2𝜆 → 0
...
21

• This means that if the budget constraint is increased (i
...
loosened) by R1, utility will increase by 0
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5𝑦𝑥 −1

• Since 𝑥 = 10; 𝑦 = 40, this means that 𝑀𝑅𝐶𝑆 = 0

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