bee1020—basicmathematicaleconomics dieterbalkenborg week14...

BEE1020 — Basic Mathematical Economics Dieter BalkenborgWeek 14, Lecture Thursday 26.01.2006 Department of EconomicsOptimization University of Exeter

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

is the work of a most perfect creator, nothing whatso-

ever takes place in the universe in which some form of

maximum or minimum does not appear.”

Leonhard Euler, 1744

optimization problems:– unconstrained (profit maximization)– constrained (utility maximization with budget constraint)“first order conditions” for optimumyield simultaneous system of equations (→Review)Lagrangian approach for constrained problems

Unconstrained optimization

Find absolute maximum of function

z = f (x, y)

i.e., find pair (x∗, y∗) such that

f (x∗, y∗) ≥ f (x, y)

for all (x, y).The following must hold: freeze the variable y at optimal valuey∗, vary only x then

f (x, y∗)

Must have maximum at x∗. Obtain first order conditions

∂z

∂x|x=x∗,y=y∗= 0

∂z

∂y |x=x∗,y=y∗= 0

2

-40

-30

-20

-10

0

-4 -22 4

y

-4-2

24

x

3

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

Π (K,L) = PQ (K,L)− rK − rL.

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...

4

Rewrite as∂Q

∂K=r

P(3)

∂Q

∂L=w

P(4)

Division yields:∂Q

∂K

/∂Q

∂L=r

P

/ wP

=r

w(5)

Marginal rate of substitution must equal ratio input prices.

5

Example:

Q (K,L) = K16L

12

P = 12, r = 1 and w = 3. Then

∂Q

∂K=

1

6K−

56L

12

∂Q

∂L=

1

2K

16L−

12

∂Q

∂K

/∂Q

∂L=

1

6K−

56L

12

/1

2K

16L−

12 =

1

3K−

56L

12K−

16L

12 =

1

3

L

K

FOC’s:1

6K−

56L

12 =

1

12K−

56L

12 =

1

2(6)

1

2K

16L−

12 =

3

12K

16L−

12 =

1

2(7)

6

FOC’s:1

6K−

56L

12 =

1

12K−

56L

12 =

1

2(8)

1

2K

16L−

12 =

3

12K

16L−

12 =

1

2(9)

Division yields1

3

L

K=

1

3L = K

Hence

K−56L

12 = K−

56K

36 = K−

13 =

1

2or

3

√1

K=

1

2⇔

1

K=

1

8⇔ K = L = 8

7

10

20

K

0 2 4 612

16 18 20 22

L

-10

-8

-6

-4

-2

0

2

4

6

8

10

12

14

16

18

20

z

8

Example: A monopolist with total cost function TC (Q) = Q2

sells his product in two different countries. When he sells QA unitsof the good in country A he will obtain the price

PA = 22− 3QA

for each unit. When he sells QB units of the good in country Bhe obtains 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)QB9

Total production costs are:

TC = (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 (10)

2QA + 10QB = 34.

linear simultaneous system

10

Simultaneous systems of equations

Linear systems

example

5x + 7y = 50 (11)

4x− 6y = −18

Using the slope-intercept form

7y = 50− 5x

y =50

7−

5

7x

4x− 6y = −18

4x + 18 = 6y

y =2

3x + 3

11

2

3

4

5

6

7

8

-2 0 2 4 6x

12

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

50

7−

5

7x = y =

2

3x + 3

50

7− 3 =

2

3x +

5

7x | × 3× 7

150− 63 = 14x + 15x

87 = 29x

x =87

29= 3

found value of x. calculate the value of y:

y =2

3× 3 + 3 = 5

solution: x∗ = 3, y∗ = 5

13

strongly recommended to check:

5× 3 + 7× 5 = 15 + 35 = 50

4× 3− 6× 5 = 12− 30 = −18

14

The method of substitution

solve one of the equations for one of the variables:

4x− 6y = −18

4x + 18 = 6y

y =4

6x + 3 =

2

3x + 3 (12)

replace y in the other equation. (brackets!), obtain equation in onevariable.

15

5x + 7y = 50

5x+ 7

(2

3x + 3

)= 50

5x +14

3x + 21 = 50

15

3x +

14

3x + 21 = 50

29

3x = 50− 21 = 29

x =3

29× 29 = 3

use (12) to find y:

y =2

3x + 3 =

2

3× 3 + 3 = 5

solution: x = 3, y = 5.

16

The method of elimination

multiply first equation by coefficient of x in second equationmultiply second equation by the coefficient of x in first equation

5x + 7y = 50 | × 4

4x− 6y = −18 | × 5

20x+ 28y = 200

20x− 30y = −90

subtract second equation from first equation

20x + 28y = 20020x− 30y = −90 |−0 + 28y − (−30y) = 200− (−90)58y = 290

y = 29058 = 5

17

use one of the original equations to find x

5x + 7y = 50

5x + 35 = 50

5x = 15

x = 3

We could have eliminated y first:

5x + 7y = 50 | × 6

4x− 6y = −18 | × 7

30x+ 42y = 300

28x− 42y = −126 | +

58x = 174

x = 3

18

Cramer’s rule

→ linear algebra. Write system of equations as[5 74 −6

] [xy

]=

[50

−18

]

where

[x1x2

]and

[50

−18

]“columns vectors”[5 74 −6

]

is the 2× 2-“matrix of coefficients”.With each 2× 2-matrix

A =

[a bc d

]

associate a new number called the determinant

detA =

∣∣∣∣a bc d

∣∣∣∣ = ad− cb

19

vertical lines, not square brackets!!!

For instance,∣∣∣∣5 74 −6

∣∣∣∣ = 5× (−6)− 4× 7 = −30− 28 = −58

Cramer’s rule for system

ax+ by = e

cx+ dy = f

or [a bb d

] [xy

]=

[ef

].

x∗ =

∣∣∣∣e bf d

∣∣∣∣∣∣∣∣a bc d

∣∣∣∣

=ed− bf

ad− bcy∗ =

∣∣∣∣a ec f

∣∣∣∣∣∣∣∣a bc d

∣∣∣∣

=ae− ce

ad− bc

20

In our example,

x∗ =

∣∣∣∣50 7

−18 −6

∣∣∣∣∣∣∣∣5 74 −6

∣∣∣∣

=50× (−6)− (−18)× 7

5× (−6)− 4× 7=−300 + 126

−30− 28=−174

−58= 3

y∗ =

∣∣∣∣5 504 −18

∣∣∣∣∣∣∣∣5 74 −6

∣∣∣∣

=5× (−18)− 4× 50

5× (−6)− 4× 7=−90− 200

−58=−290

−58= 5

Exercise:

8QA + 2QB = 22

2QA + 10QB = 34.

21

Existence and uniqueness.

linear simultaneous system of equations

ax+ by = e

cx+ dy = f

Rewrite as

y =e

b−a

bx

y =f

d−c

dx

slopes identical when ab = c

d, i.e., when determinant ad − cb iszero. If in addition intercepts are equal, both equations describethe same line.Example:

x + 2y = 3

2x + 4y = 422

has no solution: The two lines

y =3

2−

1

2x

y = 1−1

2x

are parallel

-1

0

2

-2 -1 1 2 3 4 5x

There is no common solution. Trying to find one yields a contra-

23

diction3

2−

1

2x = y = 1−

1

2x | +

1

2x

3

2= 1

24

One equation non-linear, one linear

Consider, for instance,

y2 + x− 1 = 0

y +1

2x = 1

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

1

2x = 1− y

x = 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. See example above

25

Constrained optimization

Example: Utility maximization

preferences of a consumer described by family of indifferencecurves.mathematically convenient as level curves u (x, y) =constant ofutility function.example: u (x, y) = xy. indifference curves for u = 1, u = 2 andu = 3:

0

1

2

3

4

5

1 2 3 4 5x

price of apples is p, price of oranges is q budget b.

26

budget constraint

px + qy ≤ b

27

maximize utility subject to the budget constraint! constrained

optimization problem

u (x, y) objective function

total expenditure g (x, y) = px + qy is called the constraining

function.Principles lecture budget line must be tangential to the indifferencecurvemarginal rate of substitution must equal relative price:budget line:

px + qy = b or y =b

q−p

qx

−pq = negative of relative price of apples in terms of oranges.slope of the indifference curve is negative of marginal rate of sub-

stitution∂u/∂x∂u/∂y

In optimum:28

∂u/∂x

∂u/∂y=p

q(13)

For the example u (x, y) = xy: ∂u∂x = y and ∂u∂y = x

y

x=p

q(14)

orqy = px, (15)

budget equationpx + qy = b (16)

To repeat: the constrained optimum (x∗, y∗) is the solution toa simultaneous system of two equations in two unknowns. Thefirst equation expresses that in the optimum the marginal rate ofsubstitution is equal to the relative price. The second one statesthat the consumer spends all his money in the optimum.

29

y = pqx. Substituting into budget equation gives

px+ q

(p

qx

)= px + px = 2px = b

x∗ =b

2p

y∗ =p

qx∗ =

p

q

b

2p=b

2q

When the budget is b = 100 and p = 2 and q = 5 then

x∗ =100

2× 2= 25 y =

100

2× 5= 10.

Cost minimization

dual to the utility maximization problemproduction function Q (K,L) = KL.least costly way to produce Q0 units given prices r and w

30

minimize total costsrK + wL

subject toQ0 ≤ Q (K,L) .

at cost minimum iso-cost line must be tangential to the isoquant.iso-cost line

rK + wL = constant

has slop − rwslope of isoquant is −

∂Q/∂K∂Q/∂L

In optimum:∂Q/∂K

∂Q/∂L=r

w.

31

In our example ∂Q∂K = L and ∂Q∂L = K, so

L

K=r

wor

wL = rK

so in optimum equal amounts are spend on both inputs. This isthe first equation.As the second equation we have that the firm will produce exactlyQ0 units:

Q0 = Q (K,L) .

suppose r = 2, w = 5, Q0 = 250optimum (K∗, L∗) :

L

K=

2

5or L =

2

5K

32

and250 = KL.

Substituting in the last equation L = 25K yields

250 = 2× 125 = 2× 53 =2

5K2 54 = K2 K∗ = 52 = 25

L∗ =2

5K∗ = 10.

The optimal input combination is (K∗, L∗) = (25, 10).The general problem

maximize or minimize an objective function

z = f (x, y)

subject to a constraint

g (x, y) ≤ c

where c is a constant.33

interested in case where constraint is binding in optimum, i.e.,g (x, y) = cThe optimum must then solve the two conditions

∂f/∂x

∂f/∂y=∂g/∂x

∂g/∂yg (x, y) = c

The Lagrangian approach

alternative way to derive these two conditions. The method trans-forms the constrained optimization problem into an unconstrainedoptimization problem in conjunction with a pricing problem.λ ≥ 0 is the so-called Lagrange multiplier.

Lagrangian

L (x, y) = f (x, y)− λ (g (x, y)− c)

— look for an unconstrained maximum or a minimum of this func-tion

34

— determine the Lagrange multiplier such that constraint holdswith equalitySuppose found absolute maximum (x∗, y∗) of Lagrangian. Sup-

pose g (x∗, y∗) = c.

L (x∗, y∗) = f (x∗, y∗)− λ (g (x∗, y∗)− c) = f (x∗, y∗) .

consider (x, y) with g (x, y) ≤ c.Then L (x, y) ≤ L (x∗, y∗) because (x∗, y∗) is absolute maximum.−λ (g (x, y)− c) is not positiveHence f (x, y) ≤ L (x, y) ≤ L (x∗, y∗) ≤ f (x∗, y∗)

FOC for maximum of Lagrangian:

∂L

∂x=∂f

∂x− λ

∂g

∂x= 0 or

∂f

∂x= λ

∂g

∂x(17)

∂L

∂y=∂f

∂y− λ

∂g

∂y= 0 or

∂f

∂y= λ

∂g

∂y(18)

35

divide:∂f/∂x

∂f/∂y=∂g/∂x

∂g/∂y. (19)

In addition constraint must hold in optimum

g (x, y) = c. (20)

Utility maximization again

Lagrangian

L (x, y) = u (x, y)−λ (px+ qy − b) = u (x, y) +λ (b− px + py)

The savings:s = b− px + py

future utility from saving:λs

λ is the marginal utility of saving a penny.Cost minimization again

36

The Lagrangian

L (K,L) = −rK − wL− λ (−Q (K,L) +Q0)

= (λQ (K,L)− rK − wL)− λQ0

Up to a constant the Lagrangian is profit function for output priceλ.

37