Introduction to the calculus of multi-variable functions

1. Basic Calculus (II) Recap
(for MSc & PhD Business, Management & Finance Students)
Lecturer: Farzad Javidanrad
First Draft: Sep. 2013
Revised: Sep. 2014
Multi-Variable
Functions

2. Multi-Variable Functions
• In the case of one-variable function, in the form of 𝑦 = 𝑓(𝑥) , the
variable 𝒙 is called “independent variable” and 𝒚 “dependent
variable”.
• There are many examples of the dependency of 𝑦 on 𝑥 (e.g, the state
of boiling of water depends on the amount of heat; or consumption
expenditure depends on the level of income) but the concept of
function should be understood beyond the concept of dependency.
In most of the cases, dependency is not the issue at all. The modern
concept of function is based on the idea of mapping.

3. Multi-Variable Functions
• When a painter paint a scene on a canvas s(he) uses a
correspondence rule (mapping rule): every point in three-
dimensional space (𝑅3) is corresponded (mapped) to just one and
only one point in two-dimensional space (𝑅2).
• Mathematically speaking the
function 𝑓: 𝑅3
→ 𝑅2
can
represent the type of
corresponding (mapping)
rule that the painter is
applying.

4. The Concept of Function as Mapping
• Transformation of an object is a mapping from 𝑅2 to 𝑅2;
• Mathematical operations describe a function from 𝑅2 to 𝑅
x
y
y
-xo
𝑓: 𝑅2 → 𝑅2
𝑎, 𝑏 → (𝑏, −𝑎)
(𝑎, 𝑏)
(𝑏, −𝑎)
a
b
a+boo
Figure1-6: Geometrical interpretation of
the sum operator as a function. This is a
transformation from space to .
xx
𝑔: 𝑅2 → 𝑅
𝑎, 𝑏 → 𝑎 + 𝑏

5. Multi Variables Functions
• All basic mathematical operators such as summation, subtraction,
division and multiplication introduce a function from two-
dimensional space (𝑅2) to the real number set (one-dimensional
space, 𝑅), that is:
𝑓: 𝑅2
→ 𝑅
For e.g. for division: 𝑎, 𝑏 →
𝑎
𝑏
(𝑏 ≠ 0)
• One of the important family of the multi-variable functions is the
“real (scalar) multi variables function”, which can be shown as
𝑓: 𝑅 𝑛 → 𝑅 or simply, 𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥 𝑛), where 𝑦 is the
dependent variable and 𝑥1, 𝑥2, … , 𝑥 𝑛 are independent variables.

6. Two Variables Functions
• A simple form of this function is when we have two independent
variables 𝑥, 𝑦 and one dependent variable 𝑧, in the form of 𝑧 =
𝑓(𝑥, 𝑦). This is called “two variables function” as there are two
independent variables.
• E.g. a Cobb-Douglas production function :
𝑌 = 𝑓 𝐾, 𝐿 = 𝐴𝐾 𝛼
𝐿 𝛽
Where 𝑌 is the level of production,
𝐾 and 𝐿 are the levels of capital and
labour employed for production,
respectively.
• 𝐴, 𝛼 and 𝛽 are constants of the
function.
Adoptedfrom http://en.citizendium.org/wiki/File:Cobb-Douglas_with_dimishing_returns_to_scale.png
Y
K
L

7. Two Variables Functions
• 𝑧 = 𝑓(𝑥, 𝑦) represents a functional relationship if for every ordered
pair (𝑥, 𝑦) in the domain of the function there will be one and only
one value of 𝑧 in the range of the function.
o Which graph does represent a function?
𝒙 𝟐
𝒂 𝟐
+
𝒚 𝟐
𝒃 𝟐
+
𝒛 𝟐
𝒄 𝟐
= 𝟏
Ellipsoid
Hyperboloid of Two Sheets
−
𝒙 𝟐
𝒂 𝟐
−
𝒚 𝟐
𝒃 𝟐
+
𝒛 𝟐
𝒄 𝟐
= 𝟏
Hyperbolic Paraboloid
𝒙 𝟐
𝒂 𝟐
−
𝒚 𝟐
𝒃 𝟐
=
𝒛
𝒄
Elliptic Paraboloid
𝒙 𝟐
𝒂 𝟐
+
𝒚 𝟐
𝒃 𝟐
=
𝒛
𝒄
Adoptedfromhttp://tutorial.math.lamar.edu/Classes/CalcIII/QuadricSurfaces.aspx

8. Derivative of Two Variables Functions
• Consider the function 𝑧 = 𝑓(𝑥, 𝑦); 𝑧 changes if 𝑥 or 𝑦 or both of
them change. If we control the change of 𝑦 and allow just 𝑥 to
change then the average change of 𝑧 in terms of 𝑥, is
Δ𝑧
Δ𝑥
. The limiting
state of this ratio when ∆𝑥 → 0 is what is called “partial derivative of
𝒛 in terms of 𝒙 ” and is shown by:
𝜕𝑧
𝜕𝑥
,
𝜕𝑓(𝑥,𝑦)
𝜕𝑥
, 𝑧 𝑥
′
, 𝑓𝑥
• This cutter plane shows that the
variable 𝑦 is controlled (fixed)
at 𝑦 = 1 but 𝑥 can change from
-2 to +2 and the movement is on
the curve of intersection between
The plane and the surface of the
function.
Adoptedfrom http://msemac.redwoods.edu/~darnold/math50c/matlab/pderiv/index.xhtml

9. Partial Differentiation
• If 𝑥 is controlled (fixed) and 𝑦 is allowed to change the partial
derivative of 𝒛 in terms of 𝒚 can be shown by:
𝜕𝑧
𝜕𝑦
,
𝜕𝑓(𝑥,𝑦)
𝜕𝑦
, 𝑧 𝑦
′
,𝑓𝑦
• The cutter plane shows that
𝑥 is controlled (fixed) at
𝑥 = 0 but 𝑦 can change from
-3 to +3 on the curve of intersection
between the plane and the surface
of the function.
z
y
x
Adoptedfrom http://www.uwec.edu/math/Calculus/216-Spring2007/assignments.htm

10. Partial Differentiation
• So, in general, the slope of the function 𝑧 = 𝑓(𝑥, 𝑦) on the curve of
intersection between the surface of the function and the cutting
plane parallel to x-axis at any point of the domain is:
𝜕𝑧
𝜕𝑥
= 𝑓𝑥 = lim
∆𝑥→0
𝑓 𝑥 + ∆𝑥 , 𝑦 − 𝑓(𝑥 , 𝑦)
∆𝑥
= 𝑙𝑖𝑚
ℎ→0
𝑓 𝑥 + ℎ , 𝑦 − 𝑓(𝑥 , 𝑦)
ℎ
It means when calculating
𝜕𝑧
𝜕𝑥
the
variable 𝑦 should be treated as a
constant. The same rule applies for
multi variables functions.
Adoptedfrom http://moodle.capilanou.ca/mod/book/view.php?id=19700&chapterid=240
𝒛 = 𝟏𝟎 − 𝒙 𝟐
− 𝒚 𝟐

11. Partial Differentiation
• And the slope of the function 𝑧 = 𝑓(𝑥, 𝑦) on the curve of
intersection between the surface of the function and the cutting
plane parallel to y-axis at any point of the domain is:
𝜕𝑧
𝜕𝑦
= 𝑓𝑦 = 𝑙𝑖𝑚
∆𝑦→0
𝑓 𝑥 , 𝑦 + ∆𝑦 − 𝑓(𝑥, 𝑦)
∆𝑦
= 𝑙𝑖𝑚
ℎ→0
𝑓 𝑥 , 𝑦 + ℎ − 𝑓(𝑥, 𝑦)
ℎ
It means when calculating
𝜕𝑧
𝜕𝑦
the
variable 𝑥 should be treated as a
constant. The same rule applies for
multi variables functions.
Adoptedfrom http://moodle.capilanou.ca/mod/book/view.php?id=19700&chapterid=240
𝒛 = 𝟏𝟎 − 𝒙 𝟐
− 𝒚 𝟐

12. Partial Differentiation
• To find the partial derivatives (slope of tangent lines on the surface)
at a specific point 𝑃(𝑎, 𝑏, 𝑐) we have:
•
𝜕𝑓(𝑥,𝑦)
𝜕𝑥 𝑝(𝑎,𝑏,𝑐)
= 𝑙𝑖𝑚
ℎ→0
𝑓 𝑎+ℎ , 𝑏 −𝑓(𝑎 , 𝑏)
ℎ
•
𝜕𝑓(𝑥,𝑦)
𝜕𝑦 𝑝(𝑎,𝑏,𝑐)
= 𝑙𝑖𝑚
ℎ→0
𝑓 𝑎 , 𝑏+ℎ −𝑓(𝑎 , 𝑏)
ℎ
Example:
o Find partial derivatives of 𝑧 = 10𝑥2
𝑦3
.
𝝏𝒛
𝝏𝒙
= 𝟐𝟎𝒙𝒚 𝟑 ,
𝝏𝒛
𝝏𝒚
= 𝟑𝟎𝒙 𝟐 𝒚 𝟐
(𝒂, 𝒃, 𝟎)
(𝒂, 𝒃, 𝒄)
(𝟎, 𝒃, 𝟎)
(𝒂, 𝟎, 𝟎)
Adoptedfrom http://www.solitaryroad.com/c353.html

13. Rules of Partial Differentiation
• If 𝑓(𝑥, 𝑦) and 𝑔(𝑥, 𝑦) are two differentiable functions with respect
to 𝑥 and 𝑦 ;
 𝑧 = 𝑓 𝑥, 𝑦 ± 𝑔 𝑥, 𝑦 →
𝜕𝑧
𝜕𝑥
=
𝜕𝑓
𝜕𝑥
±
𝜕𝑔
𝜕𝑥
= 𝑓𝑥 ± 𝑔 𝑥
𝜕𝑧
𝜕𝑦
=
𝜕𝑓
𝜕𝑦
±
𝜕𝑔
𝜕𝑦
= 𝑓𝑦 ± 𝑔 𝑦
 𝑧 = 𝑓 𝑥, 𝑦 × 𝑔 𝑥, 𝑦 →
𝜕𝑧
𝜕𝑥
= 𝑓𝑥 . 𝑔 + 𝑔 𝑥 . 𝑓
𝜕𝑧
𝜕𝑦
= 𝑓𝑦 . 𝑔 + 𝑔 𝑦 . 𝑓
 𝑧 =
𝑓(𝑥,𝑦)
𝑔(𝑥,𝑦)
→
𝜕𝑧
𝜕𝑥
=
𝑓𝑥 . 𝑔−𝑔 𝑥 . 𝑓
𝑔2
𝜕𝑧
𝜕𝑦
=
𝑓𝑦 . 𝑔−𝑔 𝑦 . 𝑓
𝑔2

14. Some Examples
o Find partial derivatives of the function 𝑧 = 𝑥2 − 𝑥𝑦3 − 5𝑦2.
𝜕𝑧
𝜕𝑥
= 𝟐𝒙 − 𝒚 𝟑 ,
𝜕𝑧
𝜕𝑦
= −𝟑𝒙𝒚 𝟐 − 𝟏𝟎𝒚
o Find partial derivatives of 𝑧 = 𝑥𝑦. 𝑥2 + 𝑦2 .
𝜕𝑧
𝜕𝑥
= y. 𝑥2 + 𝑦2 +
2𝑥
2 𝑥2 + 𝑦2
. 𝑥𝑦 = 𝐲. 𝒙 𝟐 + 𝒚 𝟐 +
𝒙 𝟐 𝒚
𝒙 𝟐 + 𝒚 𝟐
𝜕𝑧
𝜕𝑦
= 𝑥. 𝑥2 + 𝑦2 +
2𝑦
2 𝑥2 + 𝑦2
. 𝑥𝑦 = 𝒙. 𝒙 𝟐 + 𝒚 𝟐 +
𝒚 𝟐
𝒙
𝒙 𝟐 + 𝒚 𝟐
o Find partial derivatives of 𝑧 =
3𝑥2 𝑦2
𝑥4+𝑦4.
𝜕𝑧
𝜕𝑥
=
6𝑥𝑦2
𝑥4
+ 𝑦4
− 4𝑥3
× 3𝑥2
𝑦2
𝑥4 + 𝑦4 2
𝜕𝑧
𝜕𝑦
=
6𝑦𝑥2
𝑥4
+ 𝑦4
− 4𝑦3
× 3𝑥2
𝑦2
𝑥4 + 𝑦4 2

15. Chain Rule (Different Cases)
Case 1: If 𝑧 = 𝑓 𝑢 and 𝑢 = 𝑔(𝑥, 𝑦) then 𝑧 = 𝑓(𝑔 𝑥, 𝑦 ) and
Examples:
o Find partial derivatives of 𝑧 = 𝑒 𝑥𝑦2
.
Suppose 𝑢 = 𝑥𝑦2 , so, 𝑧 = 𝑒 𝑢 and
𝜕𝑧
𝜕𝑥
=
𝜕𝑧
𝜕𝑢
.
𝜕𝑢
𝜕𝑥
= (𝑒 𝑢). 𝑢 𝑥 = 𝒆 𝒙𝒚 𝟐
. 𝒚 𝟐
and
𝜕𝑧
𝜕𝑦
=
𝜕𝑧
𝜕𝑢
.
𝜕𝑢
𝜕𝑦
= (𝑒 𝑢). 𝑢 𝑦 = 𝒆 𝒙𝒚 𝟐
. 𝟐𝒙𝒚
𝜕𝑧
𝜕𝑥
= 𝑓′.
𝜕𝑔
𝜕𝑥
=
𝜕𝑧
𝜕𝑢
.
𝜕𝑢
𝜕𝑥
and
𝜕𝑧
𝜕𝑦
= 𝑓′.
𝜕𝑔
𝜕𝑦
=
𝜕𝑧
𝜕𝑢
.
𝜕𝑢
𝜕𝑦

16. Chain Rule (Different Cases)
o Find partial derivatives of the function 𝑧 = 𝑒
𝑥
𝑦 + cos(𝑥𝑦) .
𝜕𝑧
𝜕𝑥
=
𝟏
𝒚
𝒆
𝒙
𝒚 − 𝒚. 𝒔𝒊𝒏 𝒙𝒚 ,
𝜕𝑧
𝜕𝑦
=
−𝒙
𝒚 𝟐
𝒆
𝒙
𝒚 − 𝒙. 𝒔𝒊𝒏(𝒙𝒚)
• Case 2: If 𝑧 = 𝑓 𝑥, 𝑦 is a differentiable function of 𝑥 and 𝑦 and
these two variables are differentiable functions of 𝑟 , such that 𝑥 =
𝑔 𝑟 and 𝑦 = ℎ(𝑟) , then:
o Find partial derivatives of 𝑧 = 𝑥 − 𝑙𝑛𝑦 when 𝑥 = 𝑟 and 𝑦 = 𝑟2 − 1
𝜕𝑧
𝜕𝑟
= 1.
1
2 𝑟
−
1
𝑦
. 2𝑟 =
𝟏
𝟐 𝒓
−
𝟐𝒓
𝒓 𝟐 − 𝟏
• Can you suggest another way?
𝜕𝑧
𝜕𝑟
=
𝜕𝑧
𝜕𝑥
.
𝑑𝑥
𝑑𝑟
+
𝜕𝑧
𝜕𝑦
.
𝑑𝑦
𝑑𝑟
The same rules apply
for multi variables
functions

17. Chain Rules (Different Cases)
• Case 3: If 𝑧 = 𝑓 𝑥, 𝑦 is a differentiable function of 𝑥 and 𝑦 and
these two variables are differentiable functions of 𝑟 and 𝑠 , such
that 𝑥 = 𝑔 𝑟, 𝑠 and 𝑦 = ℎ(𝑟, 𝑠) and 𝑟 and 𝑠 are independent from
each other (
𝑑𝑟
𝑑𝑠
,
𝑑𝑠
𝑑𝑟
= 0), then:
• These derivatives are called “total derivatives of 𝒛 with respect to
𝒓 and 𝒔”.
o Find partial derivatives of 𝑧 =
3
𝑥2 − 𝑦 where 𝑥 = 𝑟2 + 𝑠2 and
𝑦 =
𝑟
𝑠
.
𝜕𝑧
𝜕𝑟
=
𝜕𝑧
𝜕𝑥
.
𝑑𝑥
𝑑𝑟
+
𝜕𝑧
𝜕𝑦
.
𝑑𝑦
𝑑𝑟
and
𝜕𝑧
𝜕𝑠
=
𝜕𝑧
𝜕𝑥
.
𝑑𝑥
𝑑𝑠
+
𝜕𝑧
𝜕𝑦
.
𝑑𝑦
𝑑𝑠

18. Implicit Differentiation
• The Chain Rule can be used for implicit differentiation even for one
variable functions:
𝐹 𝑥, 𝑦 = 0
Using the chain rule we have:
𝜕𝐹
𝜕𝑥
=
𝜕𝐹
𝜕𝑥
.
𝑑𝑥
𝑑𝑥
+
𝜕𝐹
𝜕𝑦
.
𝑑𝑦
𝑑𝑥
= 0
So,
• The same rule can be used for implicit two or multi variables functions.
For example, for an implicit function 𝐹 𝑥, 𝑦, 𝑧 = 0, we have:
As
𝒅𝒙
𝒅𝒙
= 𝟏
𝑑𝑦
𝑑𝑥
= −
𝜕𝐹
𝜕𝑥
𝜕𝐹
𝜕𝑦
= −
𝐹𝑥
𝐹𝑦
𝜕𝑧
𝜕𝑥
= −
𝜕𝐹
𝜕𝑥
𝜕𝐹
𝜕𝑧
= −
𝐹𝑥
𝐹𝑧
𝑎𝑛𝑑
𝜕𝑧
𝜕𝑦
= −
𝜕𝐹
𝜕𝑦
𝜕𝐹
𝜕𝑧
= −
𝐹𝑦
𝐹𝑧

19. Examples of Implicit Functions
o Find the slope of the tangent line on the curve of intersection
between the surface 𝑥2 + 𝑦2 + 𝑧2 = 9 and the plane 𝑦 = 2 at the
point 𝐴(1,2,2) .
As 𝑦 is fixed at 2 so, we are looking for
𝜕𝑧
𝜕𝑥
at point A :
2𝑥 + 0 + 2𝑧.
𝜕𝑧
𝜕𝑥
= 0 →
𝜕𝑧
𝜕𝑥
=
−𝑥
𝑧
= −
1
2
Or using implicit differentiation:
𝜕𝑧
𝜕𝑥
= −
𝐹𝑥
𝐹𝑧
= −
2𝑥
2𝑧
= −
𝑥
𝑧
o Find
𝜕𝑧
𝜕𝑦
for 𝑒 𝑥+𝑦+𝑧 = 𝑥2 − 2𝑦2 + 𝑧2 .
0 + 1 +
𝜕𝑧
𝜕𝑦
𝑒 𝑥+𝑦+𝑧 = 0 − 4y + 2z.
𝜕𝑧
𝜕𝑦
→
𝜕𝑧
𝜕𝑦
=
𝑒 𝑥+𝑦+𝑧 + 4𝑦
2𝑧 − 𝑒 𝑥+𝑦+𝑧
Use the implicit differentiation for this question.

20. Higher Orders Partial Derivatives
• For the function 𝑧 = 𝑓(𝑥, 𝑦) the partial derivatives
𝜕𝑧
𝜕𝑥
and
𝜕𝑧
𝜕𝑦
are in
turn functions of 𝑥 and 𝑦 , in general. So, we can think of second
partial derivatives of 𝑧 , but in this case there are three different
second derivatives:
𝑧 𝑥𝑥 = 𝑓𝑥𝑥 =
𝜕
𝜕𝑧
𝜕𝑥
𝜕𝑥
=
𝜕
𝜕𝑥
𝜕𝑧
𝜕𝑥
=
𝜕2 𝑧
𝜕𝑥2
𝑧 𝑦𝑦 = 𝑓𝑦𝑦 =
𝜕
𝜕𝑧
𝜕𝑦
𝜕𝑦
=
𝜕
𝜕𝑥
𝜕𝑧
𝜕𝑦
=
𝜕2
𝑧
𝜕𝑦2
𝑧 𝑥𝑦 = 𝑓𝑥𝑦 =
𝜕
𝜕𝑧
𝜕𝑥
𝜕𝑦
=
𝜕
𝜕𝑦
𝜕𝑧
𝜕𝑥
=
𝜕2
𝑧
𝜕𝑦. 𝜕𝑥
Second-
order direct
partial
derivatives
Second-
order cross
partial
derivative

21. The Equality of Mixed (Cross) Partial Derivatives
𝑧 𝑦𝑥 = 𝑓𝑦𝑥 =
𝜕
𝜕𝑧
𝜕𝑦
𝜕𝑥
=
𝜕
𝜕𝑥
𝜕𝑧
𝜕𝑦
=
𝜕2
𝑧
𝜕𝑥. 𝜕𝑦
• If the cross (mixed) partial derivatives 𝑓𝑥𝑦 and 𝑓𝑦𝑥 are continuous
and finite in their domain then they are equal to one another; i.e.
𝑓𝑥𝑦 = 𝑓𝑦𝑥
Or
𝜕2 𝑧
𝜕𝑦.𝜕𝑥
=
𝜕2 𝑧
𝜕𝑥.𝜕𝑦
𝑧 = 𝑓(𝑥, 𝑦)
𝜕𝑧
𝜕𝑥
=𝑓𝑥
𝜕𝑧
𝜕𝑦
=𝑓𝑦
𝑓𝑥𝑥
𝑓𝑥𝑦 = 𝑓𝑦𝑥
𝑓𝑦𝑦
Second-
order cross
partial
derivative

22. Total Differential
• The meaning of differential in multi variables scalar function is not
different with that in the one variable function. The only difference is
that the source of change in dependent variable is the change of all
independent variables., that is;
𝑧 + ∆𝑧 = 𝑓(𝑥 + ∆𝑥, 𝑦 + ∆𝑦)
Or ∆𝑧 = 𝑓 𝑥 + ∆𝑥, 𝑦 + ∆𝑦 − 𝑓(𝑥, 𝑦)
But 𝑑𝑧, which is called “total differential”
is defined as:
𝑑𝑧 =
𝜕𝑧
𝜕𝑥
. 𝑑𝑥 +
𝜕𝑧
𝜕𝑦
𝑑𝑦
Or
𝑑𝑧 = 𝑓𝑥. 𝑑𝑥 + 𝑓𝑦. 𝑑𝑦 Adoptedfrom Calculus Early Transcendental James Stewart p897

23. Total Differential
• For a multi variables scalar function the same rule applies:
𝑧 = 𝑓(𝑥1, 𝑥2, … , 𝑥 𝑛)
𝑑𝑧 =
𝜕𝑧
𝜕𝑥1
. 𝑑𝑥1 +
𝜕𝑧
𝜕𝑥2
. 𝑑𝑥2 + ⋯ +
𝜕𝑧
𝜕𝑥 𝑛
. 𝑑𝑥 𝑛
• in the case of two variables function 𝑧 = 𝑓(𝑥, 𝑦) we assumed 𝑥 and 𝑦 are
independent, but if they depend on other variables the differential of
each one of them can be treated as the total differential of a dependent
variable, that is;
𝑧 = 𝑓 𝑥, 𝑦 → 𝑑𝑧 =
𝜕𝑧
𝜕𝑥
. 𝑑𝑥 +
𝜕𝑧
𝜕𝑦
. 𝑑𝑦 𝐴
𝑥 = ℎ 𝑟, 𝑠 → 𝑑𝑥 =
𝜕𝑥
𝜕𝑟
. 𝑑𝑟 +
𝜕𝑥
𝜕𝑠
. 𝑑𝑠 𝐵
𝑦 = 𝑘 𝑟, 𝑠 → 𝑑𝑦 =
𝜕𝑦
𝜕𝑟
. 𝑑𝑟 +
𝜕𝑦
𝜕𝑠
. 𝑑𝑠 𝐶
Substituting B and C into A:
𝑑𝑧 =
𝜕𝑧
𝜕𝑥
.
𝜕𝑥
𝜕𝑟
. 𝑑𝑟 +
𝜕𝑥
𝜕𝑠
. 𝑑𝑠 +
𝜕𝑧
𝜕𝑦
.
𝜕𝑦
𝜕𝑟
. 𝑑𝑟 +
𝜕𝑦
𝜕𝑠
. 𝑑𝑠

24. Total Differential
If we are looking for total derivatives of 𝑧 with respect to 𝑟 and 𝑠,
which is introduced before as the chain rule (case 3), we need to
suppose that 𝑟 and 𝑠 are independent variables and not associated to
each other (
𝑑𝑠
𝑑𝑟
𝑜𝑟
𝑑𝑟
𝑑𝑠
= 0); then:
𝑑𝑧 =
𝜕𝑧
𝜕𝑥
.
𝜕𝑥
𝜕𝑟
+
𝜕𝑧
𝜕𝑦
.
𝜕𝑦
𝜕𝑟
. 𝑑𝑟 +
𝜕𝑧
𝜕𝑥
.
𝜕𝑥
𝜕𝑠
+
𝜕𝑧
𝜕𝑦
.
𝜕𝑦
𝜕𝑠
. 𝑑𝑠
𝜕𝑧
𝜕𝑟
=
𝜕𝑧
𝜕𝑥
.
𝑑𝑥
𝑑𝑟
+
𝜕𝑧
𝜕𝑦
.
𝑑𝑦
𝑑𝑟
and
𝜕𝑧
𝜕𝑠
=
𝜕𝑧
𝜕𝑥
.
𝑑𝑥
𝑑𝑠
+
𝜕𝑧
𝜕𝑦
.
𝑑𝑦
𝑑𝑠

25. Second Order Total Differential
• The sign of the second order total differential 𝑑2 𝑧 shows the convexity and
concavity of the surface with respect to the 𝑥𝑜𝑦 plane.
• Considering the total differential 𝑑𝑧 , the second order total differential
𝑑2 𝑧 can be obtained by applying the differential rules:
𝑑2 𝑧 = 𝑑 𝑑𝑧 = 𝑑(
𝜕𝑧
𝜕𝑥
. 𝑑𝑥 +
𝜕𝑧
𝜕𝑦
𝑑𝑦)
= 𝑑 𝑓𝑥. 𝑑𝑥 + 𝑓𝑦. 𝑑𝑦
= 𝑑𝑓𝑥. 𝑑𝑥 + 𝑓𝑥. 𝑑 𝑑𝑥 + 𝑑𝑓𝑦. 𝑑𝑦 + 𝑓𝑦. 𝑑(𝑑𝑦)
As
𝑑 𝑑𝑥 = 𝑑2
𝑥 = 0
𝑑 𝑑𝑦 = 𝑑2
𝑦 = 0
, and
𝑑𝑓𝑥 = 𝑓𝑥𝑥. 𝑑𝑥 + 𝑓𝑥𝑦. 𝑑𝑦
𝑑𝑓𝑦 = 𝑓𝑦𝑥. 𝑑𝑥 + 𝑓𝑦𝑦. 𝑑𝑦
therefore :
• Factorising 𝑑𝑥2
from the right hand side, we have:
𝑑2 𝑧 = 𝑑𝑦2. 𝑓𝑥𝑥.
𝑑𝑥
𝑑𝑦
2
+ 2𝑓𝑥𝑦.
𝑑𝑥
𝑑𝑦
+ 𝑓𝑦𝑦
𝑑2 𝑧 = 𝑓𝑥𝑥. 𝑑𝑥2 + 2𝑓𝑥𝑦. 𝑑𝑥. 𝑑𝑦 + 𝑓𝑦𝑦. 𝑑𝑦2

26. Second Order Differential
• 𝑑𝑦2
> 0 (why?); so the sign of 𝑑2
𝑧 depends on the sign of the
expression in the bracket.
• From elementary algebra we know that the quadratic form
𝑎𝑋2 + 𝑏𝑋 + 𝑐 has the same sign as the parameter 𝑎 when ∆=
𝑏2
− 4𝑎𝑐 < 0 .
• If we assume that 𝑋 =
𝑑𝑥
𝑑𝑦
and 𝑎 = 𝑓𝑥𝑥 , 𝑏 = 2𝑓𝑥𝑦 , 𝑐 = 𝑓𝑦𝑦 then
𝑑2
𝑧 = 𝑑𝑦2
. 𝑎𝑋2
+ 𝑏𝑋 + 𝑐 has the same sign as 𝑎 = 𝑓𝑥𝑥 if
2𝑓𝑥𝑦
2
− 4𝑓𝑥𝑥. 𝑓𝑦𝑦 < 0 → 𝑓𝑥𝑥. 𝑓𝑦𝑦 > 𝑓𝑥𝑦
2
So;
1. 𝑑2
𝑧 > 0 if 𝑓𝑥𝑥 > 0 and 𝑓𝑥𝑥. 𝑓𝑦𝑦 > 𝑓𝑥𝑦
2
.
2. 𝑑2 𝑧 < 0 if 𝑓𝑥𝑥 < 0 and 𝑓𝑥𝑥. 𝑓𝑦𝑦 > 𝑓𝑥𝑦
2
.
Adoptedfrom Calculus Early Transcendental James Stewart DIFFERENT PAGES

27. Optimising of Two Variables Functions
• The two variables function 𝑧 = 𝑓(𝑥, 𝑦) have a relative maximum
(relative minimum) at a point in its domain if at that point :
Note 1: If 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
< 0 , it
means the critical point is not a maximum
or a minimum but a saddle point.
(looks maximum from one axis but
minimum from another axis)
Adoptedfrom http://commons.wikimedia.org/wiki/File:Saddle_point.png
𝒛 = 𝒙 𝟐
− 𝒚 𝟐
i. 𝑓𝑥 = 0 and 𝑓𝑦 = 0 , simultaneously.
ii. 𝑓𝑥𝑥 < 0 (𝑓𝑥𝑥 > 0)
iii. 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
> 0
Sufficient Conditions
Necessary conditions for
differentiable functions

28. Optimising of Two Variables Functions
• Note 2: If 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
= 0 at the critical point further
investigation is needed to find about the nature of the point.
• Example:
o Find the local extremum of the function 𝑓(𝑥, 𝑦) = 𝑥3 − 6𝑥𝑦 + 8𝑦3,
if any.
𝑓𝑥 = 0
𝑓𝑦 = 0
→
6𝑥2
− 6𝑦 = 0
−6𝑥 + 24𝑦2
= 0
→
𝑥2
= 𝑦
−𝑥 + 4𝑦2
= 0
After solving these simultaneous equations two critical points emerge
𝑨(𝟎, 𝟎, 𝟎) and 𝑩(
𝟑 𝟏
𝟒
,
𝟑 𝟏
𝟏𝟔
,
−𝟑
𝟒
) .

29. Optimising of Two Variables Functions
Now, 𝑓𝑥𝑥 = 12𝑥 and 𝑓𝑦𝑦 = 48𝑦 and 𝑓𝑥𝑦 = 𝑓𝑦𝑥 = −6 .
So, 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
= 12𝑥. 48𝑦 − −6 2 = 576𝑥𝑦 − 36 .
At the point 𝐴 0,0,0 : 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
= −36 < 0 →
𝐴 𝐢𝐬 𝐚 𝐬𝐚𝐝𝐥𝐥𝐞 𝐩𝐨𝐢𝐧𝐭.
At the point
𝐵(
3 1
4
,
3 1
16
,
−3
4
) :𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
= 144 − 36 = 108 > 0 and
𝑓𝑥𝑥 > 0 , so this point is a local minimum.

30. The Jacobian & Hessian Determinants
• From the matrix algebra we know that for any square matrix 𝐴
if:
𝐴 = 0 ⟹ 𝐴 𝑖𝑠 𝑎 𝑠𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥,
Which means, there exists linear dependence between at least two
rows or two columns of the matrix.
And if:
𝐴 ≠ 0 ⟹ 𝐴 𝑖𝑠 𝑎 𝑛𝑜𝑛 − 𝑠𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑎𝑡𝑟𝑖𝑥,
Which means, all rows and all columns are linearly independent.
• So to test for linear dependence between the equations in a
simultaneous system the determinant of the coefficients matrix
can be used.

31. The Jacobian & Hessian Determinants
• To test for functional dependence (both linear and non-linear) between
different functions we use Jacobian Determinant shown by 𝐽 .
• The Jacobian Matrix is the matrix of all first-order partial derivatives of a
vector function 𝐹: 𝑅 𝑛 → 𝑅 𝑚, which corresponds a vector in 𝑛
dimensional space(real n-tuples) into a vector in 𝑚 dimensional space
(real m-tuples):
𝑦1 = 𝐹1(𝑥1, 𝑥2, … , 𝑥 𝑛)
𝑦2 = 𝐹2(𝑥1, 𝑥2, … , 𝑥 𝑛)
⋮
𝑦 𝑚 = 𝐹𝑚(𝑥1, 𝑥2, … , 𝑥 𝑛)
So, the Jacobian matrix of 𝐹 is:
𝐽 =
𝜕𝐹1
𝜕𝑥1
⋯
𝜕𝐹1
𝜕𝑥 𝑛
⋮ ⋱ ⋮
𝜕𝐹𝑚
𝜕𝑥1
⋯
𝜕𝐹𝑚
𝜕𝑥 𝑛
Each row is the partial
derivatives of one of the
functions (e.g. 𝐹1) with
respect to all independent
variables 𝑥1, 𝑥2, … , 𝑥 𝑛.

32. The Jacobian & Hessian Determinants
• If 𝑚 = 𝑛, the Jacobian matrix is a square matrix and its
determinant shows if there is functional dependence or
independence between the functions.
𝐽 = 0 ⟹ 𝑇ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙𝑙𝑦 𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡
This means, there is a linear or non-linear association between two
functions.
𝐽 ≠ 0 ⟹ 𝑇ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡
This means, there is no linear or non-linear association between
two functions.
Example: Use the Jacobian determinant to test the functional
dependency of the following equations:
𝑦1 = 2𝑥1 − 3𝑥2
𝑦2 = 4𝑥1
2
− 12𝑥1 𝑥2 + 9𝑥2
2

33. The Jacobian & Hessian Determinants
• The Jacobian determinant is :
𝐽 =
𝜕𝑦1
𝜕𝑥1
𝜕𝑦1
𝜕𝑥2
𝜕𝑦2
𝜕𝑥1
𝜕𝑦2
𝜕𝑥2
=
2 −3
8𝑥1 − 12𝑥2 −12𝑥1 + 18𝑥2
= 2 −12𝑥1 + 18𝑥2 − −3 8𝑥1 − 12𝑥2 = 0
• So, the functions are not independent.
• We expected such a result as we know that there is a quadratic
functional relationship between 𝑦1 and 𝑦2:
𝑦1
2
= 𝑦2

34. The Jacobian & Hessian Determinants
• Hessian Matrix is a square matrix which is composed of the second-
order partial derivatives of a real (scalar) multi variables function,
(𝑓: 𝑅 𝑛 → 𝑅). For a function 𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥 𝑛), Hessian
determinant is defined as:
𝐻 =
𝜕2
𝑓
𝜕𝑥1
2
𝜕2
𝑓
𝜕𝑥1 𝜕𝑥2
⋯
𝜕2
𝑓
𝜕𝑥1 𝜕𝑥 𝑛
𝜕2
𝑓
𝜕𝑥2 𝜕𝑥1
𝜕2
𝑓
𝜕𝑥2
2 ⋯
𝜕2
𝑓
𝜕𝑥2 𝜕𝑥 𝑛
⋮
𝜕2
𝑓
𝜕𝑥 𝑛 𝜕𝑥1
⋮
𝜕2
𝑓
𝜕𝑥 𝑛 𝜕𝑥2
…
⋮
𝜕2
𝑓
𝜕𝑥 𝑛
2
=
𝑓11 𝑓12 … 𝑓1𝑛
𝑓21 𝑓22 … 𝑓2𝑛
⋮
𝑓𝑛1
⋮
𝑓𝑛2
⋱ ⋮
… 𝑓𝑛𝑛
• In the optimisation of two variables function if the first-order
(necessary) conditions 𝑓𝑥 = 𝑓𝑦 = 0 are met, second-order
(sufficient) conditions are:
 𝑓𝑥𝑥, 𝑓𝑦𝑦 > 0 𝑓𝑜𝑟 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 and 𝑓𝑥𝑥, 𝑓𝑦𝑦 < 0 𝑓𝑜𝑟 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
 𝑓𝑥𝑥 . 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2
> 0

35. The Jacobian & Hessian Determinants
• Using the Hessian determinant, we can simply show the sufficient
conditions as:
 The optimal point is minimum if 𝐻1 > 0 and 𝐻2 > 0, because:
o 𝐻1 = 𝑓𝑥𝑥 > 0
o 𝐻2 =
𝑓𝑥𝑥 𝑓𝑥𝑦
𝑓𝑦𝑥 𝑓𝑦𝑦
= 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦
2 > 0
 And, the optimal point is maximum if 𝐻1 < 0 and 𝐻2 > 0.
• There is the same story for a multi-variable function 𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥 𝑛):
 If 𝐻1 , 𝐻2 , 𝐻3 , … , 𝐻 𝑛 > 0, the critical point is local minimum.
 If the principal minors change their signs consecutively, the critical point is
the local maximum. (e.g. in case of 𝑦 = 𝑓(𝑥1, 𝑥2, 𝑥3), 𝐻1 < 0 , 𝐻2 > 0 and
𝐻3 < 0)

36. Optimisation with a Constraint
• In reality, independent variables in a function, are not fully
independent from each other. They might be in a linear or even
non-linear relationship with one another and make a constraint in
the process of optimisation and change the result of that.
Adoptedfrom http://staff.www.ltu.se/~larserik/applmath/chap7en/part7.html
Adopted& altered from http://en.wikipedia.org/wiki/Lagrange_multiplier
𝑔 𝑥, 𝑦 = 𝑐
Linear constraintNon-linear constraint
𝒈 𝒙, 𝒚 = 𝒄

37. Optimisation with a Constraint
• In each case, the function 𝑧 = 𝑓(𝑥, 𝑦) is the target function for
optimisation, subject to a constraint 𝑔 𝑥, 𝑦 = 𝑐 (where 𝑐 is a
constant). So;
Max or Min ∶ 𝑧 = 𝑓 𝑥, 𝑦
Subject to ∶ 𝑔 𝑥, 𝑦 = 𝑐
• If the constraint function 𝑔 𝑥, 𝑦 = 𝑐 is linear (e.g. 𝑥 − 2𝑦 = −1)
one way to include the constraint into the optimisation process is to
find one variable with respect to another from the constraint
function (here; 𝑥 = 2𝑦 − 1) and put it into the target function to
make it a function with one independent variable, 𝑧 = 𝐹(𝑦), and
follow the optimisation process of two-variable function.

38. Example
• Example: Find the maximum of the function 𝑧 = 𝑥𝑦 subject to
the constraint 𝑥 + 𝑦 = 1.
From the constraint function we have 𝑦 = −𝑥 + 1 and if we
substitute this with the 𝑦 in the target function, we will have 𝑧 =
− 𝑥2 + 𝑥.
𝑑𝑧
𝑑𝑥
= 0 → −2𝑥 + 1 = 0 → 𝑥 = 0.5
Putting this into the constraint equation to find 𝑦 and both into the
target function to find 𝑧 ; the maximum point
will be 𝐴(0.5, 0.5, 0.25) .
 How do we know the point is the maximum point?

39. The Lagrange Method
• If the constraint function is non-linear the previous method
might become very complicated. Another method, which is
called “Lagrange Method” or the “Method of Lagrange
Multipliers”, can help us to find local extremum points.
• In the Lagrange method the constraint function comes into the
process of optimisation by introducing a new variable λ
(Lagrange Multiplier, Lagrange coefficient) to make the Lagrange
function 𝐿 , in the form of:
𝐿 𝑥, 𝑦, 𝜆 = 𝑓 𝑥, 𝑦 + 𝜆 . [𝑐 − 𝑔 𝑥, 𝑦 ]
• By changing 𝑥 and 𝑦 a point is moving on the surface of the
function but the movement is limited to the constraint 𝑔 𝑥, 𝑦 =
𝑐 .
• This means 𝑐 − 𝑔 𝑥, 𝑦 = 0 and 𝐿 𝑥, 𝑦, 𝜆 = 𝑓 𝑥, 𝑦 . So, the
optimisation of 𝐿 is equivalent to the optimisation of 𝑓 .

40. The Lagrange Method
• To find the extremum values we need to find the derivative of the
Lagrange function with respect to its variables and solve the
following simultaneous equations :
𝜕𝐿
𝜕𝑥
= 0 →
𝜕𝑓
𝜕𝑥
− 𝜆.
𝜕𝑔
𝜕𝑥
= 0
𝜕𝐿
𝜕𝑦
= 0 →
𝜕𝑓
𝜕𝑦
− 𝜆.
𝜕𝑔
𝜕𝑦
= 0
𝜕𝐿
𝜕𝜆
= 0 → 𝑐 − 𝑔 𝑥, 𝑦 = 0
• Solving this simultaneous equations gives us the critical values of 𝑥
and 𝑦 and a value for 𝜆 .
• 𝜆 shows the sensitivity of the target (objective)function to the
change in the constraint function.
Necessary conditions
for having extremums A

41. Sufficient Condition
• To make sure that the critical point(s) from solving the
simultaneous equations are extremum(s) we need sufficient
evidence which is the sign of second order differential of the
Lagrange function 𝑑2 𝐿 at the critical point(s).
• If 𝐿 = 𝑓 𝑥, 𝑦 + 𝜆 . [c − 𝑔 𝑥, 𝑦 ] then
𝑑𝐿 = 𝑑𝑓 − 𝑔. 𝑑𝜆 − 𝜆. 𝑑𝑔
And
𝑑2 𝐿 = 𝑑2 𝑓 − 𝑑𝑔. 𝑑𝜆 − 𝑔. 𝑑2 𝜆 − 𝑑𝜆 . 𝑑𝑔 − 𝜆 . 𝑑2 𝑔
Since:
 𝑑2 𝜆 = 0
 𝑑2
𝑓 = 𝑓𝑥𝑥. 𝑑𝑥2
+ 2𝑓𝑥𝑦. 𝑑𝑥. 𝑑𝑦 + 𝑓𝑦𝑦. 𝑑𝑦2
 𝑑𝑔 = 𝑔 𝑥. 𝑑𝑥 + 𝑔 𝑦. 𝑑𝑦
 𝑑2 𝑔 = 𝑔 𝑥𝑥. 𝑑𝑥2 + 2𝑔 𝑥𝑦. 𝑑𝑥. 𝑑𝑦 + 𝑔 𝑦𝑦. 𝑑𝑦2
, therefore

42. 𝑑2
𝐿 = 𝑓𝑥𝑥 − 𝜆. 𝑔 𝑥𝑥 . 𝑑𝑥2
+ 2 𝑓𝑥𝑦 − 𝜆. 𝑔 𝑥𝑦 . 𝑑𝑥. 𝑑𝑦 + 𝑓𝑦𝑦 − 𝜆. 𝑔 𝑦𝑦 . 𝑑𝑦2
−
2𝑔 𝑥 𝑑𝑥. 𝑑𝜆 − 2𝑔 𝑦 𝑑𝑦. 𝑑𝜆
= 𝐿 𝑥𝑥. 𝑑𝑥2 + 2𝐿 𝑥𝑦. 𝑑𝑥. 𝑑𝑦 + 𝐿 𝑦𝑦. 𝑑𝑦2 − 2𝑔 𝑥 𝑑𝑥. 𝑑𝜆 − 2𝑔 𝑦 𝑑𝑦. 𝑑𝜆
• In the matrix form we can use the bordered Hessian Matrix to represent the
above quadratic form:
𝑑2 𝐿 = 𝑑𝜆 𝑑𝑥 𝑑𝑦
0 −𝑔 𝑥 −𝑔 𝑦
−𝑔 𝑥 𝐿 𝑥𝑥 𝐿 𝑥𝑦
−𝑔 𝑦 𝐿 𝑦𝑥 𝐿 𝑦𝑦
𝑑𝜆
𝑑𝑥
𝑑𝑦
• Where the bordered Hessian matrix is:
𝐻3 =
0 −𝑔 𝑥 −𝑔 𝑦
−𝑔 𝑥 𝐿 𝑥𝑥 𝐿 𝑥𝑦
−𝑔 𝑦 𝐿 𝑦𝑥 𝐿 𝑦𝑦
or sometimes 𝐻3 =
𝐿 𝑥𝑥 𝐿 𝑥𝑦 −𝑔 𝑥
𝐿 𝑦𝑥 𝐿 𝑦𝑦 −𝑔 𝑦
−𝑔 𝑥 −𝑔 𝑦 0
Sufficient Condition

43. • In the second form, the components of vectors of the first differentials of
the variables, need to be re-arranged, i.e.:
𝑑𝑥 𝑑𝑦 𝑑𝜆
𝐿 𝑥𝑥 𝐿 𝑥𝑦 −𝑔 𝑥
𝐿 𝑦𝑥 𝐿 𝑦𝑦 −𝑔 𝑦
−𝑔 𝑥 −𝑔 𝑦 0
𝑑𝑥
𝑑𝑦
𝑑𝜆
• Note: In some books the constraint function 𝑔 enters in the Lagrange
function with a positive sign, so, the signs of the first derivatives of 𝑔
in the bordered Hessian matrix are positive, but there is no difference
between their determinants. (Based on the properties of determinant,
if just a row or just a column of a matrix multiplied by 𝑘, the
determinant of the matrix is multiplied by 𝑘. In this case, the first row
and the first column multiplied by -1, so, the determinant is multiplied
by -1x(-1)=1)
Sufficient Condition

44. So, we have a minimum if
1. 𝑑2 𝐿 > 0 (i.e. all the principle minors of the Hessian matrix should
be negative: 𝐻2 , 𝐻3 < 0 )
And a maximum if:
2. 𝑑2 𝐿 < 0 (i.e. the principle minors of the Hessian matrix change their
sign one after another: 𝐻2 > 0, 𝐻3 < 0 )
• For a multi variable function 𝑦 = 𝑓(𝑥1, 𝑥2, … , 𝑥 𝑛), The Hessian matrix
is 𝑛 × 𝑛 but the rule is the same:
• For minimum: 𝐻2 , 𝐻3 , … , 𝐻 𝑛 < 0 .
• For maximum:
Tℎ𝑒 𝑠𝑖𝑔𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑚𝑖𝑛𝑜𝑟𝑠 𝑐ℎ𝑎𝑛𝑔𝑒 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒𝑙𝑦.
Sufficient Condition

45. Example
• Find the extremums of the function 𝑓 𝑥, 𝑦 = 𝑥 − 𝑦 subject to the 𝑥2
+ 𝑦2
=
100, if any?
𝐿 𝑥, 𝑦, 𝜆 = 𝑥 − 𝑦 + 𝜆[100 − 𝑥2 − 𝑦2]
𝐿 𝑥 = 1 − 2𝜆𝑥 = 0
𝐿 𝑦 = −1 − 2𝜆𝑦 = 0
𝐿 𝜆 = 100 − 𝑥2 − 𝑦2 = 0
1
−1
=
2𝜆𝑥
2𝜆𝑦
From the first two equations 𝜆 can be omitted and we have 𝑥 = −𝑦. Substituting
this new equation into the third equation we will have:
100 −𝑦2 − 𝑦2 = 0 → 𝑦 = ±5 2
So, the critical points are A −5 2, 5 2, −10 2 and 𝐵(5 2, −5 2, 10 2) and
𝜆 = ∓
2
20
.
Without any further investigation it can be said that point A is minimum and point
𝐵 is maximum. (Why?)

46. Example
• Using hessian determinant method we have:
𝐻3 =
0 −2𝑥 −2𝑦
−2𝑥 −2𝜆 0
−2𝑦 0 −2𝜆
= 8𝜆(𝑥2 + 𝑦2)
Obviously, the sign of this determinant depends on the sign of 𝜆.
 At point A −5 2, 5 2, −10 2 , 𝜆 = −
2
20
,so, 𝐻3 <0 and the point is
minimum. ( 𝐻2 is also negative).
 At point 𝐵 5 2, −5 2, 10 2 , 𝜆 = +
2
20
, so, 𝐻3 >0 and the point is
maximum.
• If there are more than one constraint the process of optimisation is
the same but there will be more than one Lagrange multiplier.
• This case is the generalisation of the previous case and will not be
discussed here.

47. Interpretation of the Lagrange Multiplier 𝜆
• The first-order conditions in the form of the simultaneous
equations (slide 40), provides the critical (and perhaps) optimal
values of the independent variables (𝑥∗, 𝑦∗) and the corresponding
value(s) of the Lagrange multiplier (𝜆∗).
• The Lagrange multiplier shows the sensitivity of the optimal value
of the target (objective) function(𝑓∗) to the change in the constant
value of the constraint function (𝑐). It is calculated as the ratio, i.e.:
𝜆∗ =
𝜕𝑓∗ 𝑥∗, 𝑦∗
𝜕𝑐
This means if 𝜆∗ = 2, and 𝑐 increases by 1%, the value of the target
function (calculated at the optimal values 𝑥∗
and 𝑦∗
) increases 2%.
A

48. Duality in Optimisation Analysis
• Consider the process of maximisation of the target (objective)
function 𝑧 = 𝑓(𝑥, 𝑦), subject to the constraint 𝑔 = 𝑔(𝑥, 𝑦).
• As we know, the solution is the tangency point on both functions,
so, the process of optimisation can be done through different
approaches. The primal approach is what we have discussed and
done so far but the dual approach is when the constraint function
𝑔 = 𝑔(𝑥, 𝑦) is the new target function and 𝑧 = 𝑓(𝑥, 𝑦) as the new
constraint.
• The initial idea comes from the mathematical fact that if 𝑓 reaches
to its maximum at the point 𝑥 = 𝑥∗, the function −𝑓 will have a
minimum at that point.
• Therefore, instead of finding the maximum of 𝑧 = 𝑓(𝑥, 𝑦), subject
to the constraint 𝑔 = 𝑔 𝑥, 𝑦 , we can find the minimum of 𝑔 =
𝑔 𝑥, 𝑦 , subject to the constraint 𝑧 = 𝑓(𝑥, 𝑦), i.e. if we know that 𝑧
cannot be bigger than 𝑧∗what is the minimum value of 𝑔 𝑥, 𝑦 ,
which satisfies this constraint.

49. Duality in Optimisation Analysis
• Let 𝑈 = 𝑈(𝑥, 𝑦) is the utility function subject to the budget
constraint 𝑥. 𝑃𝑥 + 𝑦. 𝑃𝑦 = 𝑚.
• The Lagrange function is:
𝐿 𝑥, 𝑦, 𝜆 = 𝑈 𝑥, 𝑦 + 𝜆(𝑚 − 𝑥. 𝑃𝑥 − 𝑦. 𝑃𝑦)
The first-order conditions are:
𝐿 𝑥 = 𝑈 𝑥 − 𝜆𝑃𝑥 = 0
𝐿 𝑦 = 𝑈 𝑦 − 𝜆𝑃𝑦 = 0
𝐿 𝜆 = 𝑚 − 𝑥. 𝑃𝑥 + 𝑦. 𝑃𝑦 = 0
• The optimal value for 𝑥 and 𝑦 which shows the Marshallian demand
(consumption) function for 𝑥 and 𝑦 and the optimal value for 𝜆 are:
𝑥 𝑀 = 𝑥 𝑀 𝑃𝑥, 𝑃𝑦, 𝑚
𝑦 𝑀 = 𝑦 𝑀 𝑃𝑥, 𝑃𝑦, 𝑚
𝜆 𝑀 = 𝜆 𝑀 𝑃𝑥, 𝑃𝑦, 𝑚
B

50. Duality in Optimisation Analysis
• Substituting these solutions into the target function gives the
maximum value of the utility can be achieved by the constraint:
𝑈∗ = 𝑈∗ 𝑥 𝑀 𝑃𝑥, 𝑃𝑦, 𝑚 , 𝑦 𝑀 𝑃𝑥, 𝑃𝑦, 𝑚
We call this as the indirect utility function, as it is the maximum value
of the utility obtained at the optimal values of 𝑥 and 𝑦, but it is an
indirect function because now its values depends on the parameters
𝑃𝑥, 𝑃𝑦 and 𝑚.
• Now, the dual problem is when the expenditure on 𝑥 and 𝑦 is
minimised subject to the maintaining of a given level of utility 𝑈∗
.
So, the new Lagrange function is:
𝐿 𝑥, 𝑦, 𝜆 = 𝑥. 𝑃𝑥 + 𝑦. 𝑃𝑦 + 𝜆[𝑈∗ − 𝑈 𝑥, 𝑦 ]
The first-order conditions provide optimal solutions for 𝑥,𝑦 and 𝜆.

51. Duality in Optimisation Analysis
𝐿 𝑥 = 𝑃𝑥 − 𝜆𝑈 𝑥 = 0
𝐿 𝑦 = 𝑃𝑦 − 𝜆𝑈 𝑦 = 0
𝐿 𝜆 = 𝑈∗
− 𝑈(𝑥, 𝑦) = 0
The optimal solutions represent the demand functions for 𝑥 and 𝑦 .
𝑥 𝐻
= 𝑥 𝐻
𝑃𝑥, 𝑃𝑦, 𝑈∗
𝑦 𝐻
= 𝑦 𝐻
𝑃𝑥, 𝑃𝑦, 𝑈∗
𝜆 𝐻 = 𝜆 𝐻 𝑃𝑥, 𝑃𝑦, 𝑈∗
• The first two equations are called Hicksion demand functions.
Both simultaneous equations and give us the same results:
𝑈 𝑥
𝑃𝑥
=
𝑈 𝑦
𝑃𝑦
𝑜𝑟
𝑈 𝑥
𝑈 𝑦
=
𝑃𝑥
𝑃𝑦
So, primal and dual analysis leads us to the same conclusion. The only
difference is that:
𝜆 𝐻
=
1
𝜆 𝑀
C
B C