BEEM103OptimizationTechniquesforEconomistspeople.exeter.ac.uk/dgbalken/BEEM10309/week 02 slides.pdf · Functionsintwovariables Partialderivatives Optimization UnconstrainedOptimization

Functions in two variablesPartial derivatives

OptimizationUnconstrained Optimization

Second order conditions

BEEM103 Optimization Techniques for EconomistsMultivariate Functions

Dieter Balkenborg

Department of Economics, University of Exeter

Week 2

Balkenborg Multivariate Functions




1 Functions in two variablesExample: The cubic polynomialExample: production functionExample: profit function.Level CurvesIsoquants

2 Partial derivativesA Basic ExampleNotationA Second ExampleThe Marginal Products of Labour and Capital

3 Optimization4 Unconstrained Optimization

The first order conditionsExample 1Example 2: Maximizing profits





Example: The cubic polynomialExample: production functionExample: profit function.Level Curves

Example 3: Price Discrimination

5 Second order conditionsBalkenborg Multivariate Functions





Functions in two variables

A functionz = f (x , y)

or simplyz (x , y)

in two independent variables with one dependent variable assignsto each pair (x , y) of (decimal) numbers from a certain domain Din the two-dimensional plane a number z = f (x , y).x and y are hereby the independent variablesz is the dependent variable.






Outline

1 Functions in two variablesExample: The cubic polynomialExample: production functionExample: profit function.Level Curves

Isoquants


3 Optimization

4 Unconstrained OptimizationThe first order conditionsExample 1Example 2: Maximizing profitsExample 3: Price Discrimination

5 Second order conditions






Example: The cubic polynomial

The graph of f is the surface in 3-dimensional space consisting ofall points (x , y , f (x , y)) with (x , y) in D.

z = f (x , y) = x3 − 3x2 − y2

1

-5

0

y

1

x

00-1

2

z

5

2 3-1

-2

Exercise: Evaluate z = f (2, 1), z = f (3, 0), z = f (4,−4),z = f (4, 4)






Outline


Isoquants


3 Optimization








Example: production function

Q =6√K√L = K

16 L

12

capital K ≥ 0, labour L ≥ 0, output Q ≥ 0

010

20

y

200

1015

x

50

2

4z

6






Outline


Isoquants


3 Optimization








Example: profit function

Assume that the firm is a price taker in the product market and inboth factor markets.

P is the price of output









r the interest rate (= the price of capital)










w the wage rate (= the price of labour)










w the wage rate (= the price of labour)

total profit of this firm:

Π (K , L) = TR − TC= PQ − rK − wL= PK

16 L

12 − rK − wL






P = 12, r = 1, w = 3:

Π (K , L) = PK16 L

12 − rK − wL

= 12K16 L

12 −K − 3L

20x

10

z

0

10

-10

0

20

y

2010

0






P = 12, r = 1, w = 3:

Π (K , L) = PK16 L

12 − rK − wL

= 12K16 L

12 −K − 3L

20x

10

z

0

10

-10

0

20

y

2010

0

Profits is maximized at K = L = 8.Balkenborg Multivariate Functions





Outline


Isoquants


3 Optimization








Level Curves

The level curve of the function z = f (x , y) for the level c isthe solution set to the equation

f (x , y) = c

where c is a given constant.






Level Curves

The level curve of the function z = f (x , y) for the level c isthe solution set to the equation

f (x , y) = c

where c is a given constant.

Geometrically, a level curve is obtained by intersecting thegraph of f with a horizontal plane z = c and then projectinginto the (x , y)-plane. This is illustrated on the next page forthe cubic polynomial discussed above:






0

1

-50

2

01

x

2 3

5

-1z

y-1

-2

-11 2 3

0

2

1

0

xz-1

y

-2

compare: topographic map






Isoquants

In the case of a production function the level curves are calledisoquants. An isoquant shows for a given output levelcapital-labour combinations which yield the same output.

0

x

2020

y100

0z

6

010

20

20

10

x

y

z0






Finally, the linear function

z = 3x + 4y

has the graph and the level curves:

420

x

0

4

y

20

20z10

30

z

0

2

4

y

x

0 2 4

The level curves of a linear function form a family of parallel lines:

c = 3x + 4y 4y = c − 3x y =c

4− 34x

slope − 34 , variable intercept c4 .Balkenborg Multivariate Functions





Exercise: Describe the isoquant of the production function

Q = KL

for the quantity Q = 4.Exercise: Describe the isoquant of the production function

Q =√KL

for the quantity Q = 2.






Remark: The exercises illustrate the following general principle: Ifh (z) is an increasing (or decreasing) function in one variable, thenthe composite function h (f (x , y)) has the same level curves asthe given function f (x , y) . However, they correspond to differentlevels.

0

04

2

4

yx

20

20

z 10

0

04

2

4

yx

20

z 2

4

Q = KL Q =√KL





A Basic ExampleNotationA Second ExampleThe Marginal Products of Labour and Capital

Objectives for the week

Functions in two independent variables.

The lecture should enable you for instance to calculate themarginal product of labour.








Level curves ←→ indifference curves or isoquants









Level curves ←→ indifference curves or isoquants

Partial differentiation ←→ partial analysis in economics







Outline


Isoquants


3 Optimization








Partial Derivatives: A Basic Example

Exercise: What is the derivative of

z (x) = a3x2

with respect to x when a is a given constant?

Exercise: What is the derivative of

z (y) = y3b2

with respect to y when b is a given constant?






Outline


Isoquants


3 Optimization








Partial derivatives: Notation

Consider function z = f (x , y). Fix y = y0, vary only x :z = F (x) = f (x , y0).

The derivative of this function F (x) is called the partialderivative of f with respect to x and denoted by

∂z

∂x |y=y0=dF

dx









∂z

∂x |y=y0=dF

dx

Notation: “d”=“dee”, “δ”=“delta”, “∂”=“del”









∂z

∂x |y=y0=dF

dx

Notation: “d”=“dee”, “δ”=“delta”, “∂”=“del”

It suffices to think of y and all expressions containing only yas exogenously fixed constants. We can then use the familiarrules for differentiating functions in one variable in order toobtain ∂z

∂x.






Partial derivatives: Notation continued

Other common notations for partial derivatives are ∂f∂x, ∂f

∂yor

fx , fy orf′x , f

′y .








∂yor

fx , fy orf′x , f

′y .

Consider again function z = f (x , y). Fix y = y0, vary only x :z = F (x) = f (x , y0). Then evaluate at x = x0








∂yor

fx , fy orf′x , f

′y .

Consider again function z = f (x , y). Fix y = y0, vary only x :z = F (x) = f (x , y0). Then evaluate at x = x0

The derivative of this function F (x) at x = x0 is calledthe partial derivative of f with respect to x and denoted

by∂z

∂x |x=x0,y=y0=dF

dx |x=x0=dF

dx(x0)






Partial derivatives: The example continued

Example: Letz (x , y) = y3x2

Then

∂z

∂x= y32x = 2y3x

∂z

∂y= 3y2x2






Outline


Isoquants


3 Optimization








Partial derivatives: A second example

Example: Letz = x3 + x2y2 + y4.

Setting e.g. y = y0 = 1 we obtain z = x3 + x2 + 1 and hence

∂z

∂x |y=1= 3x2 + 2x

Evaluating at x = x0 = 1 we then obtain:

∂z

∂x |x=1,y=1= 5






Partial derivatives: A second example

For fixed, but arbitrary, y we obtain

∂z

∂x= 3x2 + 2xy2

as follows: We can differentiate the sum x3+x2y2+y4 withrespect to x term-by-term. Differentiating x3 yields 3x2,differentiating x2y2 yields 2xy2 because we think now of y2 as a

constant andd(ax 2)dx

= 2ax holds for any constant a. Finally, thederivative of any constant term is zero, so the derivative of y4 withrespect to x is zero.Similarly considering x as fixed and y variable we obtain

∂z

∂y= 2x2y + 4y3






Outline


Isoquants


3 Optimization








The Marginal Products of Labour and Capital

Example: The partial derivatives ∂q∂Kand ∂q

∂Lof a production

function q = f (K , L) are called the marginal product of capitaland (respectively) labour. They describe approximately by howmuch output increases if the input of capital (respectively labour)is increased by a small unit.Fix K = 64, then q = K

16 L

12 = 2L

12 which has the graph

0 1 2 3 4 50

1

2

3

4

x

Q






The Marginal Products

This graph is obtained from the graph of the function in twovariables by intersecting the latter with a vertical plane parallel toL-q-axes.

1020

z0246

y

20

x

15100 50

The partial derivatives ∂q∂Kand ∂q

∂Ldescribe geometrically the slope

of the function in the K - and, respectively, the L- direction.Balkenborg Multivariate Functions





Diminishing productivity of labour:

The more labour is used, the less is the increase in output whenone more unit of labour is employed. Algebraically:

∂q

∂L=1

2K

16 L−

12 =

1

2

6√K

2√L> 0,

∂2q

∂L2=

∂

∂L

(∂q

∂L

)= −1

4K

16 L−

32 = −1

4

6√K

2√L3< 0,

Exercise: Find the partial derivatives of

z =(x2 + 2x

) (y3 − y2

)+ 10x + 3y





Optimization

“Since the fabric of the universe is most perfect, and

is the work of a most perfect creator, nothing whatsoever

takes place in the universe in which some form of

maximum or minimum does not appear.”

Leonhard Euler, 1744





Objectives

Subject: Optimization of multivariate functions





Objectives


Two basic types of problems:





Objectives



1 unconstrained (e.g. profit maximization)





Objectives



1 unconstrained (e.g. profit maximization)2 constrained





The first order conditionsExample 1Example 2: Maximizing profitsExample 3: Price Discrimination

Unconstrained Optimization

-4-2

-4

-50

-40

-2

y x

-30

z

-20

-10

000

242

4






Unconstrained Optimization

Objective: Find (absolute) maximum of function

z = f (x , y)

i.e., find 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) .

4

y

2

4

-15

-10z-5

0

x

200

0

42

yz0

2

x4






Outline


Isoquants


3 Optimization








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 firstorder conditions

∂z

∂x |x=x ∗,y=y ∗= 0

∂z

∂y |x=x ∗,y=y ∗= 0






Outline


Isoquants


3 Optimization








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)






Outline


Isoquants


3 Optimization








Maximizing profits

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

Π (K , L) = PQ (K , L)− rK − wL.FOC for profit 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 firm could produce ∂Q∂Kunits of output more. The

revenue would increase by P ∂Q∂K, the cost by r and so profit would

increase. Thus we cannot have a profit 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=

r

P(3)

∂Q

∂L=

w

P(4)

Division yields:

MRS = − dLdK

=∂Q

∂K

/∂Q

∂L=r

P

/ wP=r

w(5)

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






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

/∂Q∂Lunits of

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

would save r on capital and spend ∂Q∂K

/∂Q∂L× w on more labour

while still making the same revenue. Profit would increase unless

r ≤ ∂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 firm optimizes profits.






Example

Q (K , L) = K16 L

12

∂Q

∂K=1

6K

16−1L

12

∂Q

∂L=1

2K

16 L

12−1

∂Q

∂K=1

6K−

56 L

12

∂Q

∂L=1

2K

16 L−

12

FOC:

1

6K−

56 L

12 =

r

P

1

2K

16 L−

12 =

w

P∂Q

∂K

/∂Q

∂L=

1

3K−

56− 1

6 L12−(− 1

2 ) =r

w

∂Q

∂K

/∂Q

∂L=

1

3

L

K=r

w






Example

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

1

6K−

56 L

12 =

1

12

1

2K

16 L−

12 =

3

12

2K−56 L

12 = 1 (*) 2K

16 L−

12 = 1

1

3

L

K=

1

3=⇒ 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






20

15

x

105

0

10

z0

-10

20

10

2015

y5

0






Outline


Isoquants


3 Optimization








Example 3: Price Discrimination

A monopolist with total cost function TC (Q) = Q2 sells hisproduct in two different 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 profits?






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

Profit:

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






Profit:

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




Second order conditions: A peak

critical point of function z = f (x , y) = −x2 − y2: (0, 0)

-4

-30

-20

0

z

-10

x y

0 2 4-2 -4

4





Second order conditions: A trough

critical point of function z = f (x , y) = +x2 + y2: (0, 0)

-4

-4

-2

x

-2

y

0

02

z

30

20

10

402

4





Second order conditions: A saddle point

critical point of function z = f (x , y) = +x2 − y2: (0, 0)

100

-10-5

50

10

z5

0

y

0

-50

0 -5 -10

105

x

-100





Second order conditions: A monkey saddle

critical point of function z = f (x , y) = yx2 − y3: (0, 0)

This and more complex possibilities will be ignored.Balkenborg Multivariate Functions




The Hessian matrix

[∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

]





Theorem

Suppose that the function z = f (x , y) has a critical point at(x∗, y ∗). If the determinant of the Hessian detH =∣∣∣∣∣

∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

∣∣∣∣∣=

∂2z

∂x2∂2z

∂y2− ∂2z

∂x∂y

∂2z

∂y∂x=

∂2z

∂x2∂2z

∂y2−(

∂2z

∂x∂y

)2

is negative at (x∗, y ∗) then (x∗, y ∗) is a saddle point.If this determinant is positive at (x∗, y ∗) then (x∗, y ∗) is a peak ora trough. In this case the signs of ∂2z

∂x 2 |(x ∗,y ∗) and∂2z∂y 2 |(x ∗,y ∗) are the

same. If both signs are positive, then (x∗, y ∗) is a trough. If bothsigns are negative, then (x∗, y ∗) is a peak.

Nothing can be said if the determinant is zero.





Theorem

Suppose that the function z = f (x , y) has a critical point at(x∗, y ∗). If the determinant of the Hessian detH =∣∣∣∣∣

∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

∣∣∣∣∣=

∂2z

∂x2∂2z

∂y2− ∂2z

∂x∂y

∂2z

∂y∂x=

∂2z

∂x2∂2z

∂y2−(

∂2z

∂x∂y

)2

is negative at (x∗, y ∗) then (x∗, y ∗) is a saddle point.If this determinant is positive at (x∗, y ∗) then (x∗, y ∗) is a peak ora trough. In this case the signs of ∂2z

∂x 2 |(x ∗,y ∗) and∂2z∂y 2 |(x ∗,y ∗) are the

same. If both signs are positive, then (x∗, y ∗) is a trough. If bothsigns are negative, then (x∗, y ∗) is a peak.

Nothing can be said if the determinant is zero.Notice that detH > 0 implies sign ∂2z

∂x 2= sign ∂2z

∂y 2





detH

< 0: saddle point

> 0:

{∂2z∂x 2> 0: trough

∂2z∂x 2< 0: peak





Example

z = x2 − y2. The partial derivatives are ∂z∂x= 2x and ∂z

∂y= −2y .

Clearly, (0, 0) is the only critical point. The determinant of theHessian is

detH =

∣∣∣∣∣

∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

∣∣∣∣∣=

∣∣∣∣2 00 −2

∣∣∣∣ = (2) (−2)− 0× 0 = −4 < 0

Hence the function has a saddle point at (0, 0).





Example

z = −x2 − y2. The partial derivatives are ∂z∂x= −2x and

∂z∂y= −2y . Clearly, (0, 0) is the only critical point. The

determinant of the Hessian is

detH =

∣∣∣∣∣

∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

∣∣∣∣∣=

∣∣∣∣−2 00 −2

∣∣∣∣ = 4 > 0

and ∂2z∂x 2< 0. Hence the function has a peak at (0, 0).





Example

z = −x2 + 52xy − y2

∂z

∂x= −2x + 5

2y

∂z

∂y=

5

2x − 2y

(0, 0) critical point.





Determinant of the Hessian is

detH =

∣∣∣∣∣

∂2z∂x 2

∂2z∂y ∂x

∂2z∂x∂y

∂2z∂y 2

∣∣∣∣∣=

∣∣∣∣−2 5

252 −2

∣∣∣∣ = (−2) (−2)−(5

2

)2

= −94< 0

Hence (0, 0) saddle point although both ∂2z∂x 2

and ∂2z∂y 2

negative.





-2

z -50

-100

x y

0-4-2

-40224

4 0

z

4

2

y

x-2

0

-4

-224 0 -4


Documents

BEEM103OptimizationTechniquesforEconomistspeople.exeter.ac.uk/dgbalken/BEEM10309/week 02 slides.pdf · Functionsintwovariables Partialderivatives Optimization UnconstrainedOptimization