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EECM 3714
Lecture 8: Unit 8

Multivariate Optimisation I
Renshaw, Ch
...
g
...
2)
If all three variables are raised to the power 1, and if there are no cross-products, then the
resulting graph is called a plane (e
...
Fig 14
...
These slices are
called sections (e
...
Fig 14
...
4)
These sections are known as iso-sections (if z is constant: iso-𝑧 section, if 𝑥 constant, then iso-x
section, if y constant, then iso-y section)
Can represent these functions with lines/curves in 2D space

FIRST ORDER PARTIAL DERIVATIVES
• Suppose 𝑧 = 𝑓(𝑥, 𝑦), then:
•

•

𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑦

= 𝑓𝑥 , slope of surface in direction of x (y is treated as a constant)

= 𝑓𝑦 , slope of surface in direction of y (x is treated as a constant)
𝜕𝑧

• Suppose that 𝑧 = 𝑔 𝑥1 , 𝑥2 , … , 𝑥𝑛 , then
= 𝑓𝑘 , slope of surface in direction of 𝑥𝑘 (other 𝑥𝑛−1
𝜕𝑥𝑘
independent variables treated as constants)
• Note the change in notation:
• For univariate functions, the derivative is

𝑑𝑦
𝑑𝑥

• For multivariate functions, the partial derivative is

𝜕𝑦
𝜕𝑥

• When partially differentiating, all variables except the one that you are differentiating with
respect to are treated as constants (or kept constant)

SECOND ORDER PARTIAL DERIVATIVES
• If 𝑧 = 𝑓(𝑥, 𝑦), then first-order partial derivatives are

𝜕𝑧
𝜕𝑥

= 𝑓𝑥 and

𝜕𝑧
𝜕𝑦

= 𝑓𝑦

• The second-order partial derivatives are then
𝜕2 𝑧
𝜕𝑥 2

=

𝜕2 𝑧
𝑓𝑥𝑥 , 2
𝜕𝑦

=

𝜕2 𝑧
𝑓𝑦𝑦 ,
𝜕𝑥𝜕𝑦

=

𝜕2 𝑧
𝑓𝑥𝑦 ,
𝜕𝑦𝜕𝑥

= 𝑓𝑦𝑥
...
e
...

2
...
𝑧 = 𝑥 0
...
7

• Also do examples 14
...
6 (p
...
𝑧 = 𝑥 2 + 3𝑥𝑦 + 𝑦 2 + 2
•

𝜕𝑧
𝜕𝑥

•

𝜕2 𝑧
𝜕𝑥 2

•

𝜕2 𝑧
𝜕𝑥𝜕𝑦

= 2𝑥 + 3𝑦;
=

𝜕 𝜕𝑧
𝜕𝑥 𝜕𝑥

=

𝜕𝑧
𝜕𝑦

= 3𝑥 + 2𝑦

= 2;

𝜕 𝜕𝑧
𝜕𝑥 𝜕𝑦

3
...
3 𝑦 0
...
𝑧 = 𝑎 ln 𝑥 + 𝑏 ln 𝑦
•

𝜕𝑧
𝜕𝑥

𝑎
𝑥

•

𝜕2 𝑧
𝜕𝑥 2

•

𝜕2 𝑧
𝜕𝑥𝜕𝑦

= ;
=

𝜕𝑧
𝜕𝑦

=

𝜕 𝜕𝑧
𝜕𝑥 𝜕𝑥

=

𝜕 𝜕𝑧
𝜕𝑥 𝜕𝑦

𝜕2 𝑧
−2
−a𝑥 ; 2
𝜕𝑦

=

𝜕2 𝑧
0;
𝜕𝑦𝜕𝑥

𝜕𝑧
𝜕𝑥

•

𝜕𝑧
𝜕𝑥

⟹ 𝑦 0
...
7 ×
𝑥 0
...
7 ×
𝜕𝑥
𝜕𝑥
−0
...
3𝑥

• ∴

𝜕𝑧
𝜕𝑥

= 0
...
7 𝑦 0
...
3 𝑦 0
...
3 is the multiplicative constant

= −𝑏𝑦 −2 •

𝜕𝑧
𝜕𝑦

= 𝑥 0
...
7𝑦 −0
...
7 ⟹

= 0
...
3 𝑦 −0
...
3 ×

SOLUTION, EXAMPLE 3
•

𝜕2 𝑧
𝜕𝑥 2

=

𝜕 𝜕𝑧
𝜕𝑥 𝜕𝑥

=

𝜕
𝜕𝑥

0
...
7 0
...
3𝑦 0
...
3𝑦
𝜕2 𝑧
𝜕𝑥 2

𝜕2 𝑧
𝜕𝑥 2

𝜕2 𝑧
𝜕𝑦 2

0
...
7

= 0
...
7 × −0
...
7

= −0
...
7 𝑦 0
...
7𝑥 0
...
3

• ⟹ 0
...
3 is the multiplicative constant
•

𝜕2 𝑧
𝜕𝑦 2

• ⟹
• ∴

𝜕

= 0
...
3 × 𝜕𝑦 𝑦 −0
...
7𝑥 0
...
3𝑦 −1
...
21𝑥 0
...
3

•

𝜕2 𝑧
𝜕𝑥𝜕𝑦

𝜕

= 𝜕𝑥

𝜕𝑧
𝜕𝑦

𝜕

= 𝜕𝑥 0
...
3 𝑦 −0
...
7𝑦 −0
...
7𝑦 −0
...
3

𝜕2 𝑧

• ⟹ 𝜕𝑥𝜕𝑦 = 0
...
3 × 0
...
7
• ∴
•

𝜕2 𝑧
𝜕𝑥𝜕𝑦

𝜕2 𝑧
𝜕𝑦𝜕𝑥

= 0
...
7 𝑦 −0
...
3𝑥 −0
...
7

• ⟹ 0
...
7 is the multiplicative constant
•

𝜕2 𝑧
𝜕𝑦𝜕𝑥

• ⟹
• ∴

𝜕

= 0
...
7 × 𝜕𝑦 𝑦 0
...
3𝑥 −0
...
7𝑦 −0
...
21𝑥 −0
...
3

ECONOMIC APPLICATION 1: PRODUCTION FUNCTIONS
• Relates inputs (capital and labour) to output (production)
• 𝑞 = 𝑓(𝐾, 𝐿)

• We’ll assume that the production function is smooth, continuous and differentiable (at least
twice)
𝜕𝑞
𝑞
, average product of labour = 𝐴𝑃𝐿 =
𝜕𝐿
𝐿
𝜕𝑞
𝑞
= , average product of capital = 𝐴𝑃𝐾 =
𝜕𝐾
𝐾

• Marginal product of labour = 𝑀𝑃𝐿 =
• Marginal product of capital = 𝑀𝑃𝐾

DIMINISHING MP & ELASTICITY
• Law of diminishing returns: as more units of one input is used, while keeping the other input
fixed, the increase in output becomes smaller and smaller
• Confirm: negative second partial derivatives (or MP curves have negative slopes), i
...

• Diminishing returns to labour:

𝜕2 𝑞
𝜕𝐿2

• Diminishing returns to capital:

𝜕2 𝑞
𝜕𝐾 2

• Labour elasticity of output = 𝑒 𝐿 =

=

𝜕 𝜕𝑞
𝜕𝐿 𝜕𝐿

<0

𝜕 𝜕𝑞
<0
𝜕𝐾 𝜕𝐾
𝑀𝑃𝐿
; capital elasticity
𝐴𝑃𝐿

=

of output = 𝑒 𝐾 =

𝑀𝑃𝐾
𝐴𝑃𝐾

COMMON FUNCTIONAL FORMS OF PRODUCTION FUNCTIONS
• You must know how to find MPL; MPK; APL; APK for all of these functions (hint: for CES, you’ll
need the chain rule)
• Cobb-Douglas: 𝑞 = 𝐴𝐾 𝛼 𝐿𝛽
𝛽

• Constant elasticity of substitution (CES): 𝑞 = 𝐴[𝛼𝐾 + 1 − 𝛼

𝛽 1ൗ𝛽
𝐿 ]

• Linear: 𝑞 = 𝑎𝐾 + 𝑏𝐿
• Log: 𝑞 = 𝑎 ln 𝐾 + 𝑏 ln 𝐿
Another type of commonly used production function is the fixed proportions production function
𝑞 = min(𝑎𝐾, 𝑏𝐿), but this function is not differentiable

ISOQUANTS AND THE MRTS
• Note that the graph of the production function q = f(K; L) is a surface in the 𝑞 − 𝐾 − 𝐿 space (e
...

Fig 14
...
g
...
15)
– has a negative slope
...
g
...
17)
• Note that 𝑀𝑃𝐿 = 𝐴𝑃𝐿 at the point where AP reaches a maximum (Fig 14
...
r
...
𝑥 = 𝑒 𝑥 =

𝑀𝑈𝑥
,
𝐴𝑈𝑥

elasticity of utility w
...
t
...


INDIFFERENCE CURVES & THE MRCS
• Note that the graph of the utility function 𝑈 = 𝑓(𝑥, 𝑦) is a surface in the 𝑢 − 𝑥 − 𝑦 space (e
...
Fig
14
...
g
...
14
...
Find 𝑑𝑥 ?
• Define z, so that 𝑧 = 0 ⟹ 𝑓(𝑥, 𝑦) = 𝑧
𝜕𝑧

𝜕𝑧

• The total differential of this function is 𝑑𝑧 = 𝜕𝑥 𝑑𝑥 + 𝜕𝑦 𝑑𝑦
𝜕𝑧

𝜕𝑧

• But 𝑧 = 0 which is a constant, so 𝑑𝑧 = 0 ⟹ − 𝜕𝑥 𝑑𝑥 = 𝜕𝑦 𝑑𝑦
• Solving for

𝑑𝑦 𝑑𝑦
:
𝑑𝑥 𝑑𝑥

=−

𝜕𝑧ൗ
𝜕𝑥
𝜕𝑧ൗ
𝜕𝑦

• We can use implicit differentiation to find the slope of iso-z sections (i
...
isoquants and indifference curves)

EXAMPLE 1, PRODUCTION FUNCTION
A firm’s production function is 𝑞 = 𝐾 0
...
5
• Find 𝑀𝑃𝐾 and 𝑀𝑃𝐿
...

• Show that 𝑀𝑅𝑇𝑆 = −

𝑀𝑃𝐿
𝑀𝑃𝐾

SOLUTION, PRODUCTION FUNCTION
• 𝑞 = 𝐾 0
...
5
• 𝑀𝑃𝐿 =

𝜕𝑞
𝜕𝐿

= 0
...
5 𝐿−0
...
5 is a multiplicative constant and differentiate 𝐿0
...
5𝐾 −0
...
5

• Note: 𝐿0
...
5
• Isoquant if 𝑞 = 100 → 100 = 𝐾 0
...
5

• Make K the subject of the formula
• Make the power of K equal to 1 by squaring both sides of the equation
• 1002 = 𝐾 0
...
5

2

• ⟹ 10000 = 𝐾𝐿
• ∴𝐾=

10000
𝐿

= 10000𝐿−1

SOLUTION, PRODUCTION FUNCTION 3
• Implicit differentiation to find the slope of the isoquant
• 100 = 𝐾 0
...
5 , where 100 = 𝑞
𝜕𝑞 Τ𝜕𝐿
𝜕𝑞 Τ𝜕𝐾

•

𝑑𝐾
−
𝑑𝐿

=

•

𝑑𝐾
𝑑𝐿
𝑑𝐾
𝑑𝐿

0
...
5 𝐿−0
...
5𝐾 −0
...
5
𝐾
− = 𝑀𝑅𝑇𝑆
𝐿

•

=
=

⟹

𝑑𝐾
𝑑𝐿

=

𝜕𝑞 Τ𝜕𝐿
− Τ
𝜕𝑞 𝜕𝐾

• So, for the Cobb-Douglas production function, the slope of the isoquant depends only on capital intensity

• Show that 𝑀𝑅𝑇𝑆 = −

𝑀𝑃𝐿
𝑀𝑃𝐾

𝜕𝑞
= 0
...
5 𝐿−0
...
5𝐾 −0
...
5
𝜕𝐾
𝑀𝑃𝐿
0
...
5 𝐿−0
...
5𝐾−0
...
5
𝐿
𝑀𝑃𝐿
𝐾
∴
= − = 𝑀𝑅𝑇𝑆
𝑀𝑃𝐾
𝐿

• 𝑀𝑃𝐿 =
•
•
•

−𝑀𝑅𝑇𝑆

EXAMPLE 2, UTILITY FUNCTION
• An individual’s utility function is 𝑈 = 𝑥 + 1

0
...
5

1
...


Find the indifference curve for 𝑈 = 6

3
...


Show that 𝑀𝑅𝐶𝑆 = −

𝑀𝑈𝑥
𝑀𝑈𝑦

SOLUTION, UTILITY FUNCTION 1
• 𝑈 = 𝑥+1

0
...
5

• 𝑀𝑈𝑥 and 𝑀𝑈𝑦 :
• 𝑀𝑈𝑥 =

𝜕𝑈
𝜕𝑥

= 0
...
5

0
...
5

is a multiplicative constant and differentiate 𝑥 + 1

= 0
...
5

0
...
5

−0
...
5

SOLUTION, UTILITY FUNCTION 2
• Indifference curve if 𝑈 = 6
• 6= 𝑥+1

0
...
5

• To make y the subject of the formula, make the power of y equal to 1 by squaring both sides of
the equation
•

62

=

0
...
5

2

⟹ 36 = 𝑥 + 1 𝑦 + 2

−2

• Implicit differentiation to find the slope of the indifference curve:
• For 6 = 𝑥 + 1
• −
•
•

𝑑𝑦
𝑑𝑥

𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥

=
=

=

𝜕𝑢Τ𝜕𝑥
𝜕𝑢Τ𝜕𝑦

0
...
5 ,

𝑦+2

𝑑𝑦
𝑑𝑥

=−

where 𝑈 = 6

𝜕𝑢Τ𝜕𝑥
𝜕𝑢Τ𝜕𝑦

0
...
5 𝑦+2 0
...
5 𝑥+1 0
...
5
𝑦+2
−
𝑥+1

SOLUTION, UTILITY FUNCTION
• Show that 𝑀𝑅𝐶𝑆 = −
• 𝑀𝑈𝑥 =
• 𝑀𝑈𝑦 =
• ⟹

𝑀𝑈𝑥
𝑀𝑈𝑦

• ∴−

𝜕𝑈
𝜕𝑥
𝜕𝑈
𝜕𝑦

=

𝑀𝑈𝑥
𝑀𝑈𝑦

𝑀𝑈𝑥
𝑀𝑈𝑦

= 0
...
5

= 0
...
5

𝑦+2

0
...
5 𝑦+2 0
...
5 𝑥+1 0
...
5

=−

𝑦+2
𝑥+1

= 𝑀𝑅𝐶𝑆

𝑦+2

=

0
...
5

𝑦+2
𝑥+1

= −𝑀𝑅𝐶𝑆

FIRST-ORDER & SECOND-ORDER CONDITIONS
• Suppose that 𝑧 = 𝑓(𝑥, 𝑦)
• To find the minimum and/or maximum values of z:

• First-order (necessary) condition:

𝜕𝑧
𝜕𝑥

=

𝜕𝑧
𝜕𝑦

=0

• Second-order (sufficient) condition:
• Maximum:

𝜕2 𝑧
𝜕𝑥 2

• Minimum:

𝜕2 𝑧
𝜕𝑥 2

< 0,

𝜕2 𝑧
𝜕𝑦 2

> 0,

𝜕2 𝑧
𝜕𝑦 2

< 0,

𝜕2 𝑧
𝜕𝑥𝜕𝑦

> 0,

𝜕2 𝑧
𝜕𝑥𝜕𝑦

×

𝜕2 𝑧
𝜕𝑦𝜕𝑥

×

𝜕2 𝑧
𝜕𝑦𝜕𝑥

<

𝜕2 𝑧
𝜕𝑥 2

𝜕2 𝑧
𝜕𝑦 2

×

<

𝜕2 𝑧
𝜕𝑥 2

𝜕2 𝑧
× 2
𝜕𝑦

EXAMPLE
• Suppose that 𝑧 = 2𝑥 + 𝑦 − 𝑥 2 + 𝑥𝑦 − 𝑦 2
• Find the values of 𝑥, 𝑦 that minimize/maximize z
...
547-8)

EXAMPLE (COBB-DOUGLAS PRODUCTION FUNCTION)
Suppose that a perfectly competitive firm can sell its output at 𝑅20, while its costs are 4𝐾 +
5𝐿 and its production function is 𝑞 = 𝐾 0
...
5
...
Show that these levels of 𝐿, 𝐾 satisfy the FOC and SOC for profit maximization
...

• 𝜋 = 𝑅 − 𝐶 = 20𝑞 − 4𝐾 + 5𝐿 = 20 𝐾 0
...
5 − 4𝐾 − 5𝐿
• FOC:

•
•

𝜕𝜋
𝜕𝐿
𝜕𝜋
𝜕𝐾

𝜕𝜋
𝜕𝐿

=

𝜕𝜋
𝜕𝐾

=0

= 10𝐾 0
...
5 − 5 = 0

[1]

= 8𝐾 −0
...
5 − 4 = 0

[2]

• Now, 10𝐾 0
...
5 = 5 [1a] and 8𝐾 −0
...
5 = 4 [2a]
•

10𝐾 0
...
5
[1a]/[2a]: −0
...
5
8𝐾
𝐿

=

5
4

SOLUTION, PROFIT MAXIMISATION
• 1
...
25 → 𝐾 = 𝐿
• Plug into [1a]: 10𝐾 0
...
5 = 5 ⟹ 10𝐾 −0
...
1 =

5
10

• Raise both sides to the power −10 =
• 𝐾

−0
...
4 𝐿−1
...
002 < 0 (substitute K=L=1024)
2
𝜕𝐿
𝜕2 𝜋
= −4
...
6 𝐿0
...
002 < 0 (substitute K=L=1024)
2
𝜕𝐾
𝜕2 𝜋
𝜕2 𝜋
=
= 4𝐾 −0
...
5 = 0
...
000004,

𝜕2 𝜋
𝜕𝐿𝜕𝐾

×

𝜕2 𝜋
𝜕𝐾𝜕𝐿

= 0
...
000004

• Therefore, profits are maximised if 𝐾 = 1024; 𝐿 = 1024, because FOC and SOC are met

• Note: for very small numbers, can use scientific notation
▪ 0
...
8 × 10−1
▪ 0
...
8 × 10−3
▪ 0
...
8 × 10−6

PRICE DISCRIMINATION
• Price discrimination: sell same product at different prices to different customers (e
...
household
and industrial; domestic and foreign, etc
...
e
...
e
...
e
...
g
...
The
inverse demand functions for the household and industrial markets are 𝑝1 = 500 − 𝑞1 and 𝑝2 =

360 − 1
...
Find the values of 𝑞1
and 𝑞2 (as well as 𝑝1 and 𝑝2 ) that will maximise profit for the monopolist
...


• Next, suppose that the monopolist does not price discriminate
...
If you were to advise the monopolist, would you advise for or against
price discrimination? Explain
...
5𝑞22 , 𝐶 = 50000 + 20 𝑞1 + 𝑞2
• 𝜋 = 𝑅1 + 𝑅2 − 𝐶 = 500𝑞1 − 𝑞12 + 360𝑞2 − 1
...
5𝑞22 − 50000
• FOC:

𝜕𝜋
𝜕𝑞1

=

𝜕𝜋
𝜕𝑞2

=0

•

𝜕𝜋
𝜕𝑞1

= 480 − 2𝑞1 = 0, ∴ 𝑞1 = 240

•

𝜕𝜋
𝜕𝑞2

= 340 − 3𝑞2 = 0, ∴ 𝑞2 = 340Τ3 = 113
...
333, because FOC and SOC met

PROFITS WITHOUT PRICE DISCRIMINATION
• Without price discrimination, 𝑝1 = 𝑝2 = 𝑝 and 𝑞 = 𝑞1 + 𝑞2
• Rewrite the inverse demand functions so that 𝑞1 , 𝑞2 are the subjects:
720−2𝑝
3
720−2𝑝
3

• 𝑞1 = 500 − 𝑝 and 1
...
t
...
p: 𝑝 =
= 444 − 0
...
6𝑞2 ⟹ 𝑀𝑅 =
= 444 − 1
...
2𝑞 = 20, ∴ 𝑞 = 353
...
67, profit without price discrimination = 24906
...

• Advice = price discriminate (greater profit)

PROFIT MAX BY A 2-PRODUCT FIRM
• Suppose a firm produces two products, then:
• 𝑅 = 𝑅1 + 𝑅2 = 𝑝1 𝑞1 + 𝑝2 𝑞2 , where 𝑝1 = 𝑓 𝑞1 , 𝑝2 = 𝑔 𝑞2

• 𝜋 = 𝑅 − 𝐶 = 𝑝1 𝑞1 + 𝑝2 𝑞2 − 𝐶, 𝐶 = ℎ(𝑞1 , 𝑞2 )
• FOC:

𝜕𝜋
𝜕𝑞1

=

𝜕𝜋
𝜕𝑞2

=0

• Or FOC: 𝑀𝑅1 = 𝑀𝐶1 , 𝑀𝑅2 = 𝑀𝐶2
•

𝜕2 𝜋
SOC: 2
𝜕𝑞1

• Or

𝜕2 𝜋
< 0, 2
𝜕𝑞2
𝜕𝑀𝑅1
SOC:
𝜕𝑞1

𝜕2 𝜋
𝜕2 𝜋
𝜕2 𝜋
< 0, 2 × 2 >
𝜕𝑞1
𝜕𝑞2
𝜕𝑞1 𝜕𝑞2
𝜕𝑀𝐶1 𝜕𝑀𝑅2
𝜕𝑀𝐶2
<
,
<
𝜕𝑞1
𝜕𝑞2
𝜕𝑞2

×

𝜕2 𝜋
𝜕𝑞2 𝜕𝑞1

EXAMPLE: PROFIT MAX BY A 2-PRODUCT FIRM
A monopolist produces two goods, A and B
...
Total cost is 𝐶 = 𝑞𝑎2 + 3𝑞𝑎 𝑞𝑏 + 𝑞𝑏2
...
Show that the first-order and second-order conditions
for profit maximization are met at these levels of 𝑞 and 𝑝
...

• 𝜋 = 𝑅𝑎 + 𝑅𝑏 − 𝐶 = 50𝑞𝑎 − 𝑞𝑎2 + 95𝑞𝑏 − 3𝑞𝑏2 − 𝑞𝑎2 − 3𝑞𝑎 𝑞𝑏 − 𝑞𝑏2

• 𝜋 = 50𝑞𝑎 + 95𝑞𝑏 − 2𝑞𝑎2 − 4𝑞𝑏2 − 3𝑞𝑎 𝑞𝑏
• FOC:
•
•
•
•
•
•

𝜕𝜋
𝜕𝑞𝑎
𝜕𝜋
𝜕𝑞𝑏

𝜕𝜋
𝜕𝑞𝑎

=

𝜕𝜋
𝜕𝑞𝑏

=0

= 50 − 4𝑞𝑎 − 3𝑞𝑏 = 0

[1]

= 95 − 3𝑞𝑎 − 8𝑞𝑏 = 0

[2]

3 × [1] and 4 × [2]
150 − 12𝑞𝑎 − 9𝑞𝑏 = 0
[1a]
380 − 12𝑞𝑎 − 32𝑞𝑏 = 0
[2a]
[1a] - [2a]: 150 − 12𝑞𝑎 − 9𝑞𝑏 − 380 + 12𝑞𝑎 + 32𝑞𝑏 = 0

• ⟹ −230 = −23𝑞𝑏 , ∴ 𝑞𝑏 = 10 ⟹ 𝑞𝑎 = 5
• 𝑝𝑎 = 45, 𝑝𝑏 = 65

SOLUTION: PROFIT MAXIMISATION BY A 2-PRODUCT FIRM
•

𝜕2 𝜋
SOC: 2
𝜕𝑞1

•

𝜕2 𝜋
2
𝜕𝑞𝑎

=

𝜕
𝜕𝑞𝑎

50 − 4𝑞𝑎 − 3𝑞𝑏 = −4 < 0

•

𝜕2 𝜋
𝜕𝑞𝑏2

=

𝜕
𝜕𝑞𝑏

95 − 3𝑞𝑎 − 8𝑞𝑏 = −8 < 0

•

𝜕2 𝜋
𝜕𝑞𝑎 𝜕𝑞𝑏

𝜕2 𝜋
×
𝜕𝑞𝑏 𝜕𝑞𝑎

•

𝜕2 𝜋
2
𝜕𝑞𝑎

𝜕2 𝜋
𝜕𝑞𝑏2

•

𝜕2 𝜋
𝜕𝑞𝑎 𝜕𝑞𝑏

×

<

𝜕2 𝜋
0, 2
𝜕𝑞2

< 0,

𝜕2 𝜋
𝜕𝑞12

×

𝜕2 𝜋
𝜕𝑞22

>

𝜕2 𝜋
𝜕𝑞1 𝜕𝑞2

×

𝜕2 𝜋
𝜕𝑞2 𝜕𝑞1

= −3

= −4 × −8 = 32

𝜕2 𝜋
×
𝜕𝑞𝑏 𝜕𝑞𝑎

= −3

2

= 9 → 9 < 32

• Therefore, profits are maximised if 𝑞𝑎 = 5, 𝑞𝑏 = 10, because FOC and SOC met

COST MINIMISATION, MULTI-PLANT FIRM
• In this application, firm produces one product, but at plants at more than one location (many
large firms do this)
• Total production cost is sum of all individual plants’ cost functions

• To minimise cost:
• FOC: Set all first partial derivatives of total cost function jointly equal to zero
• This implies that marginal cost must be equalised across all plants for the firm to minimise its total cost
• SOC: all second partial derivatives must be positive, while product of second partial derivatives must be
greater than product of cross partial derivatives

• For 2-plant firm (with cost function 𝐶 = 𝑓 𝑞1 , 𝑞2 , where 𝑞1 and 𝑞2 refer to the quantities
produced at each plant), these conditions are then
• FOC:

•

𝜕𝐶
𝜕𝑞1

𝜕2 𝐶
SOC: 2
𝜕𝑞1

=

𝜕𝐶
𝜕𝑞2

<

𝜕2 𝐶
0, 2
𝜕𝑞2

=0

< 0,

𝜕2 𝐶
𝜕𝑞12

×

𝜕2 𝐶
𝜕𝑞22

>

𝜕2 𝐶
𝜕𝑞1 𝜕𝑞2

𝜕2 𝐶
×
𝜕𝑞2 𝜕𝑞1

EXAMPLE, COST MINIMISATION, MULTI-PLANT FIRM
A firm produces computer monitors at two separate plants in Bloemfontein and
Kimberley
...
05𝑞12

− 10𝑞1 , 𝐶2 =

1
400 − 𝑞2 +
𝑞23 ,
147

while total cost is 𝐶 = 𝐶1 + 𝐶2
...


SOLUTION, COST MINIMISATION, MULTI-PLANT FIRM
• 𝐶 = 𝐶1 + 𝐶2 ⟹ 𝐶 = 400 + 0
...
05𝑞12 − 10𝑞1 − 𝑞2 +
• FOC:

𝜕𝐶
𝜕𝑞1

=

𝜕𝐶
𝜕𝑞2

1
𝑞23
147

=0

•

𝜕𝐶
𝜕𝑞1

= 0
...
1𝑞1 = 10 ⟹ 𝑞1 = 100

•

𝜕𝐶
𝜕𝑞2

= −1 +

3
𝑞22
147

3
𝑞22
147

=0⟶

= 1 ⟹ 𝑞22 = 49, → 𝑞2 = ± 49 = ±7

• But 𝑞2 ≥ 0, ∴ 𝑞2 = 7

•

𝜕2 𝐶
SOC: 2
𝜕𝑞1

•

𝜕2 𝐶
𝜕𝑞12

=

𝜕
𝜕𝐶
𝜕𝑞1 𝜕𝑞1

= 0
...
1 ×

𝜕2 𝐶
𝜕𝑞2 𝜕𝑞1

𝜕
𝜕𝐶
𝜕𝑞2 𝜕𝑞1
42
147

=

=0

42
1470

=0×0= 0<

42
1470

• Cost minimised at 𝑞1 = 100, 𝑞2 = 7, because FOC and SOC met



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