BEE1024 Mathematics for Economistspeople.exeter.ac.uk/dgbalken/ME08/week2.pdf · Introduction Unconstrained Optimization Review: Simultaneous systems of equations The Tangent Plane

IntroductionUnconstrained Optimization

Review: Simultaneous systems of equationsThe Tangent Plane

BEE1024 Mathematics for EconomistsMultivariate Functions

Juliette Stephenson and Amr (Miro) AlgarhiAuthor: Dieter Balkenborg

Department of Economics, University of Exeter

Week 2

Balkenborg Multivariate Functions



1 Introduction

2 Unconstrained OptimizationThe �rst order conditionsExample 1Example 2: Maximizing pro�tsExample 3: Price Discrimination

3 Review: Simultaneous systems of equationsUsing the slope-intercept formThe method of substitutionThe method of eliminationCramer�s ruleExistence and uniquenessOne equation non-linear, one linearTwo non-linear equations




4 The Tangent PlaneThe Total Di¤erentialThe slope of level curvesThe Chain Rule




Optimization 1

�Since the fabric of the universe is most perfect, andis the work of a most perfect creator, nothing whatsoevertakes place in the universe in which some form ofmaximum or minimum does not appear.�Leonhard Euler, 1744




Objectives

Subject: Optimization of multivariate functions

Two basic types of problems:

1 unconstrained (e.g. pro�t maximization)2 constrained (e.g. utility maximization with budget constraint)

This week:

1 ��rst order conditions� for optimum2 Simultaneous system of equations (!Review)3 the tangent plane / the marginal rate of substitution

Next weeks:

1 Lagrangian approach for constrained problems2 2nd order conditions




Objectives


Two basic types of problems:

1 unconstrained (e.g. pro�t maximization)2 constrained (e.g. utility maximization with budget constraint)

This week:


Next weeks:





Objectives


Two basic types of problems:1 unconstrained (e.g. pro�t maximization)

2 constrained (e.g. utility maximization with budget constraint)

This week:


Next weeks:





Objectives


Two basic types of problems:1 unconstrained (e.g. pro�t maximization)2 constrained (e.g. utility maximization with budget constraint)

This week:


Next weeks:





Objectives



This week:


Next weeks:





Objectives



This week:1 ��rst order conditions� for optimum

2 Simultaneous system of equations (!Review)3 the tangent plane / the marginal rate of substitution

Next weeks:





Objectives



This week:1 ��rst order conditions� for optimum2 Simultaneous system of equations (!Review)

3 the tangent plane / the marginal rate of substitution

Next weeks:





Objectives



This week:1 ��rst order conditions� for optimum2 Simultaneous system of equations (!Review)3 the tangent plane / the marginal rate of substitution

Next weeks:





Objectives




Next weeks:





Objectives




Next weeks:1 Lagrangian approach for constrained problems

2 2nd order conditions




Objectives




Next weeks:1 Lagrangian approach for constrained problems2 2nd order conditions




The �rst order conditionsExample 1Example 2: Maximizing pro�tsExample 3: Price Discrimination

Unconstrained Optimization

40

30

20

10

04 2 2 4

y42

24x





Unconstrained Optimization

Objective: Find (absolute) maximum of function

z = f (x , y)

i.e., �nd a pair (x�, y �) such that

f (x�, y �) � f (x , y)

holds for all pairs (x , y).Hereby both pairs of numbers (x�, y �) and (x , y) must be in the domain ofthe function.

For an absolute minimum require f (x�, y �) � f (x , y).f (x , y) is called the �objective function�.





Example

The function

z = f (x , y) = � (x � 2)2 � (y � 3)2

has a maximum at (x�, y �) = (2, 3) .

15

10

5

01 2 3 4 5

y

12

34

5

x1

23

45

y

4

5

x





Outline1 Introduction2 Unconstrained Optimization



4 The Tangent PlaneThe Total Di¤erentialThe slope of level curvesThe Chain Rule Balkenborg Multivariate Functions




First order conditions

The following must hold: freeze the variable y at the optimal valuey �, vary only x then the function in one variable

F (x) = f (x , y �)

must have maximum at x�: dFdx (x�) = 0. Thus we obtain the �rst

order conditions

∂z∂x jx=x �,y=y �

= 0∂z∂y jx=x �,y=y �

= 0












Example 1

z = f (x , y) = � (x � 2)2 � (y � 3)2

∂z∂x

= �2 (x � 2)� (+1) = 0 ∂z∂y= �2 (y � 3)� (+1) = 0

x� = 2 y � = 3

The maximum (at least the only critical or stationary point) is at(x�, y �) = (2, 3)












Maximizing pro�ts

Production function Q (K , L).r interest ratew wage rateP price of outputpro�t

Π (K , L) = PQ (K , L)� rK � wL.FOC for pro�t maximum:

∂Π∂K

= P∂Q∂K

� r = 0 (1)

∂Π∂L

= P∂Q∂L� w = 0 (2)





Intuition: Suppose P ∂Q∂K � r > 0. By using one unit of capital

more the �rm could produce ∂Q∂K units of output more. The

revenue would increase by P ∂Q∂K , the cost by r and so pro�t would

increase. Thus we cannot have a pro�t optimum. If P ∂Q∂K � r < 0

it would symmetrically pay to reduce capital input. HenceP ∂Q

∂K � r = 0 must hold in optimum.





Rewrite FOC as

∂Q∂K

=rP

(3)

∂Q∂L

=wP

(4)

Division yields:

MRS = � dLdK

=∂Q∂K

�∂Q∂L

=rP

. wP=rw

(5)

The marginal rate of substitution must equal the ratio of the inputprices!





Why? If the �rm uses one unit of capital less and ∂Q∂K

.∂Q∂L units of

labour more, output remains (approximately) the same. The �rm

would save r on capital and spend ∂Q∂K

.∂Q∂L � w on more labour

while still making the same revenue. Pro�t would increase unlessr � ∂Q

∂K

.∂Q∂L � w . As symmetric argument interchanging the role

of capital and labour shows that r � ∂Q∂K

.∂Q∂L � w must hold if

the �rm optimizes pro�ts.





Example

Q (K , L) = K16 L

12

∂Q∂K

=16K

16�1L

12

∂Q∂L

=12K

16 L

12�1

∂Q∂K

=16K�

56 L

12

∂Q∂L

=12K

16 L�

12

FOC:16K�

56 L

12 =

rP

12K

16 L�

12 =

wP

∂Q∂K

�∂Q∂L

=13K�

56�

16 L

12�(�

12 ) =

rw

∂Q∂K

�∂Q∂L

=13LK=rw





Example

P = 12, r = 1 and w = 3; FOC:

16K�

56 L

12 =

112

12K

16 L�

12 =

312

2K�56 L

12 = 1 (*) 2K

16 L�

12 = 1

13LK

=13=) K = L

Substituting L = K into (*) we get

2K�56K

12 = 2K�

26 = 2K�

13 = 1 =) 2 = K

13 =) K � = L� = 23 = 8

Q� = K16 L

12 = 8

16+

12 = 8

46 = 8

23 = 22 = 4

Π� = 12Q� � 1K � � 3L� = 48� 8� 24 = 16





20K

0 515 20

L

10

5

0

5

10

15

20












Example 3: Price Discrimination

A monopolist with total cost function TC (Q) = Q2 sells hisproduct in two di¤erent countries. When he sells QA units of thegood in country A he will obtain the price

PA = 22� 3QA

for each unit. When he sells QB units of the good in country B heobtains the price

PB = 34� 4QB .How much should the monopolist sell in the two countries in orderto maximize pro�ts?





Solution

Total revenue in country A:

TRA = PAQA = (22� 3QA)QA

Total revenue in country B:

TRB = PBQB = (34� 4QB )QB

Total production costs are:

TC = (QA +QB )2

Pro�t:

Π (QA,QB ) = (22� 3QA)QA + (34� 4QB )QB � (QA +QB )2





Pro�t:

Π (QA,QB ) = (22� 3QA)QA + (34� 4QB )QB � (QA +QB )2

FOC:

∂Π∂QA

= �3QA + (22� 3QA)� 2 (QA +QB ) = 22� 8QA � 2QB = 0

∂Π∂QB

= �4QB + (34� 4QB )� 2 (QA +QB ) = 34� 2QA � 10QB= 0

or

8QA + 2QB = 22 (6)

2QA + 10QB = 34.

linear simultaneous systemBalkenborg Multivariate Functions



Using the slope-intercept formThe method of substitutionThe method of eliminationCramer�s ruleExistence and uniquenessOne equation non-linear, one linearTwo non-linear equations

Review: Simultaneous systems of equations

Methods:

1 Using the slope-intercept form

2 The method of substitution3 The method of elimination4 Cramer�s rule






Methods:

1 Using the slope-intercept form2 The method of substitution

3 The method of elimination4 Cramer�s rule






Methods:

1 Using the slope-intercept form2 The method of substitution3 The method of elimination

4 Cramer�s rule






Methods:

1 Using the slope-intercept form2 The method of substitution3 The method of elimination4 Cramer�s rule












Using the slope-intercept form

Example:

5x + 7y = 50 (7)

4x � 6y = �18

slope-intercept form:

7y = 50� 5x

y =507� 57x

4x � 6y = �184x + 18 = 6y

y =23x + 3





2

3

4

5

6

7

8

2 0 2 4 6x





At intersection point two linear functions must have same y -value.Hence

507� 57x = y =

23x + 3

507� 3 =

23x +

57x j � 3� 7

150� 63 = 14x + 15x

87 = 29x

x =8729= 3

Found value of x . Calculate the value of y :

y =23� 3+ 3 = 5





5x + 7y = 50

4x � 6y = �18

solution: x� = 3, y � = 5It is strongly recommended to check, in particular in exams:

5� 3+ 7� 5 = 15+ 35 = 50

4� 3� 6� 5 = 12� 30 = �18












The method of substitution

Solve one of the equations for one of the variables:

4x � 6y = �184x + 18 = 6y

y =46x + 3 =

23x + 3 (8)

Replace y in the other equation, obtain equation in one variable.Do not forget to put brackets around the expression which replacesy !





5x + 7y = 50

5x + 7�23x + 3

�= 50

5x +143x + 21 = 50

153x +

143x + 21 = 50

293x = 50� 21 = 29

x =329� 29 = 3

Use (8) to �nd y :

y =23x + 3 =

23� 3+ 3 = 5












The method of elimination

Multiply �rst equation by coe¢ cient of x in second equation.Multiply second equation by the coe¢ cient of x in �rst equation.

5x + 7y = 50 j � 44x � 6y = �18 j � 5

20x + 28y = 200

20x � 30y = �90





Subtract second equation from �rst equation

20x + 28y = 20020x � 30y = �90 j�0+ 28y � (�30y) = 200� (�90)58y = 290y = 290

58 = 5

Use one of the original equations to �nd x

5x + 7y = 50 5x + 35 = 50 5x = 15 x = 3












Cramer�s Rule

! linear algebra. Write system of equations as�5 74 �6

� �xy

�=

�50

�18

�

where�xy

�and

�50

�18

�are �column vectors�

�5 74 �6

�is the 2� 2��matrix of coe¢ cients�.





The determinant

With each 2� 2-matrix

A =�a bc d

�associate a new number called the determinant

detA =

�� a bc d

�� = ad � cbvertical lines, not square brackets!!!For instance,�� 5 7

4 �6

�� = 5� (�6)� 4� 7 = �30� 28 = �58Balkenborg Multivariate Functions




Cramer�s rule

The linear system

ax + by = e

cx + dy = f

or �a bc d

� �xy

�=

�ef

�.

has the solution

x� =

�� e bf d

�� a bc d

�� =ed � bfad � bc y � =

�� a ec f

�� a bc d

�� =af � cead � bc





�5 74 �6

� �xy

�=

�50

�18

�

x� =

�� 50 7�18 �6

�� 5 74 �6

�� =50� (�6)� (�18)� 75� (�6)� 4� 7 =

�300+ 126�30� 28 =

�174�58 = 3

y � =

�� 5 504 �18

�� 5 74 �6

�� =5� (�18)� 4� 505� (�6)� 4� 7 =

�90� 200�58 =

�290�58 = 5





�5 74 �6

� �xy

�=

�50

�18

�

x� =

�� 50 7�18 �6

�� 5 74 �6

�� =�300+ 126�30� 28 =

�174�58 = 3

y � =

�� 5 504 �18

�� 5 74 �6

�� =�90� 200�58 =

�290�58 = 5





Exercise

8QA + 2QB = 22

2QA + 10QB = 34.












Existence and uniqueness

linear simultaneous system of equations

ax + by = e

cx + dy = f

Rewrite as

y =eb� abx

y =fd� cdx

Slopes are identical when ab =

cd , i.e., when the determinant

ad � cb is zero. If in addition intercepts are equal, both equationsdescribe the same line.





Example

x + 2y = 3 2x + 4y = 4

has no solution: The two lines

y =32� 12x y = 1� 1

2x

are parallel

1

0

2

2 1 1 2 3 4 5x





There is no common solution. Trying to �nd one yields acontradiction

32� 12x = y = 1� 1

2x

��+12x32= 1

Summary: The determinant is non-zero if and only if the systemhas a unique solution. If the determinant is zero, there are eitherzero or in�nitely many solutions.












One equation non-linear, one linear

Consider, for instance,

y2 + x � 1 = 0 y +12x = 1

In our example it is convenient to solve second equation for x :

12x = 1� yx = 2� 2y

y2 + (2� 2y)� 1 = y2 � 2y + 1 = (y � 1)2 = 0

So the unique solution is y � = 1 and x� = 2� 2� 1 = 0.












Two non-linear equations

No general method.




The Total Di¤erentialThe slope of level curvesThe Chain Rule

The Tangent Plane

For a function y = f (x) in one variable the derivative dfdx jx=x0 at

x0 is the slope of the tangent to the graph of f through the point(x0, f (x0)).

1

0

1

2

3

4

0 .2 0 .4 0 .6 0 .8 1 1 .2 1 .4 1 .6 1 .8 2x

The tangent of f (x) = x2 atx = 1.





Proposition: This tangent is the graph of the linear function

y = l (x) = f (x0) +dfdx jx=x0

(x � x0) .

Proof: This is so because l (x0) = f (x0) , so the graph of l goesthrough the point (x0, f (x0)). Moreover,

y = l (x) =�f (x0)�

dfdx0 jx=x0

x0

�+dfdx jx=x0

x ,

so l (x) is indeed a linear function in x with intercept�f (x0)� df

dx0 jx=x0x0�and the �right� slope df

dx jx=x0 .





The graph of a function in one variable is a curve and its tangentis a line. By analogy, the graph of a function z = f (x , y) in twovariables is a surface and its tangent at a point (x0, y0) is a plane.

0

1

2

3

4

5

1 2 3 4

1 2 3 4





Non-vertical planes are the graphs of linear functionsz = ax + by + c . The tangent plane of the function z = f (x , y)in (x0, y0) is, correspondingly, the graph of the linear function

z = l (x , y) = f (x0, y0)+∂f∂x jx=x0,y=y0

(x � x0)+∂f∂y jx=x0,y=y0

(y � y0)

(9)since the graph of this linear function contains the point(x0, y0, f (x0, y0)) and has the right slopes in the x- andy -directions.





Exercise: This is the example shown in the above graph. Thetangent plane of z = f (x , y) =

px +

py at the point (1, 1) is

obtained as follows: We have

∂z∂x=121px

and∂z∂y=121py

and hence

∂z∂x jx=1,y=1

=12

and∂z∂y jx=1,y=1

=12.

Since f (1, 1) = 2 the tangent plane is the graph of the linearfunction

z = l (x , y) = 2+12(x � 1) + 1

2(y � 1) = 1+ 1

2x +

12y












The Total Di¤erential

Using the abbreviations dx = x � x0, dy = y � y0 and dz = z � z0the formula for the tangent can be neatly rewritten as

dz =∂z∂xdx +

∂z∂ydy (10)

an expression which is called the total di¤erential.












The slope of level curvesThe marginal rate of substitution

xy = c

1

2

3

4

5

1 2 3 4 5x





Suppose (x0, y0) is a point on the level curve f (x , y) = c . Thenthe slope of tangent to the level curve in this point in thex-y -plane is

�MRS = dydx jx=x0,y=y0

= � ∂z∂x jx=x0,y=y0

�∂z∂y jx=x0,y=y0

(11)

provided ∂z∂y jx=x0,y=y0

6= 0. This is known as the rule for implicitdi¤erentiation: If the function y = y (x) is implicitly de�ned by theequation f (x , y (x)) = c , then its derivative dy

dx is given by theabove formula.





Remark: A quick way to memorize the formula (11) is to use thetotal di¤erential as follows: In order to stay on the level curve onemust have dz = z � z0 = 0, so

0 = dz =∂z∂xdx +

∂z∂ydy

hence

dy = ��

∂z∂x

�∂z∂y

�dx

ordydx= � ∂z

∂x

�∂z∂y.





Example: A linear function

z = ax + by + c

has the partial derivatives ∂z∂x = a and

∂z∂y = b. According to the

above formula its level curves have the slope

� ∂z∂x

�∂z∂y= � a

b.

Indeed, its level curves are the solutions to the equations

l = ax + by + c

with some constant l . Solving the equation for y we obtain

y = � abx +

l � cb

which is a linear function with slope � ab .





Example: An isoquant of the production function Q = K16 L

12 has

according to the above formula the slope

dKdL

= � ∂Q∂L

�∂Q∂K

= ��12K

16 L�

12

��16K�

56 L

12

�= �6

2K

16K

56 L�

12 L�

12 = �3K

L.

in the point (K , L).Indeed, an isoquant is the set of solutions to the equationq̄ = K

16 L

12 for �xed q̄. Solving for K we see that the isoquant is

the graph of the function

K =�q̄L�

12

�6= q̄6L�3.





Di¤erentiation yieldsdKdL

= �3q̄6L�4

and since (K , L) is on the isoquant

dKdL

= �3�K

16 L

12

�6L�4 = �3K

L.





Economically, � dKdL is the marginal rate of substitution of labour

for capital: If labour is reduced by one unit, how much morecapital is (approximately) needed to produce the same output?Just reducing labour by one unit reduces output by the marginalproduct of labour ∂q

∂L . Each additional unit of capital produces∂q∂K

�the marginal product of capital �more units. Therefore, if nowcapital input is increased by ∂q

∂L

.∂q∂K units, output changes overall

by

�∂q∂L+

�∂q∂L

�∂q∂K

�∂q∂K

= 0.





Remark: The implicit function theorem states that if∂z∂y jx=x0,y=y0

6= 0 for some point (x0, y0) on a level curvef (x , y) = c , then one can �nd a function y = g (x) de�ned nearx = x0 such that the level curve is locally the graph of thisfunction, i.e. one has f (x , g (x)) = c for x near x0.Exercise: For the production function

Q =pKL

calculate the marginal rate of substitution for K = 3 and L = 4.












The Chain rule

Suppose that z = f (x , y) is a function in two variables x , y .Suppose that x depends on the variable t via x = g (t) and that ydepends on the variable t via y = h (t). Then we can de�ne thecomposite function in one variable z = F (t) = f (x (t) , y (t))which has t as the independent and z as the dependent variable.The derivative of this function, if it exists, is calculated as

dzdt=

∂z∂xdxdt+

∂z∂ydydt.

Notice that this is just the total di¤erential divided by dt.





Example: Suppose z = xy , x = t3 + 2t, y = t2 � t. Then,according to the chain rule,

dzdt

=∂z∂xdxdt+

∂z∂ydydt= y �

�3t2 + 2

�+ x � (2t � 1)

=�t2 � t

� �3t2 + 2

�+ (2t � 1)

�t3 + 2t

�Since z = xy =

�t3 + 2t

� �t2 � t

�we get the same result using

the product rule:

dzdt=�3t2 + 2

� �t2 � t

�+�t3 + 2t

�(2t � 1) .


Documents

BEE1024 Mathematics for Economistspeople.exeter.ac.uk/dgbalken/ME08/week2.pdf · Introduction Unconstrained Optimization Review: Simultaneous systems of equations The Tangent Plane