lect 9 calculus 4

8/12/2019 Lect 9 Calculus 4

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Econ1050

Calculus 4 - Lagrange

Lecture 9


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This lecture (Differentiation)

Constrained Optimisation

first order conditions (Lagrange method)

second order conditions (Algebraic method)

meaning of Lagrange multiplier

second order conditions (Matrix method)


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What does Optimisation mean?

Finding point(s) at which

utility or profit is maximised or

costs are minimised

constrained by

technology available

competitors activity

legal restrictions onpollution

money (budget)


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Constrained Optimisation

Maximise

Subject to mpyxp

),( yxu

x

x

y

u

y

Find x* and y*


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Example 1

A firm has 3 factories each producing thesame item. Let x, y and z represent therespective output quantities from the 3factories in order to fill an order for 2000 unitsin total. Cost functions are

C1(x) = 200 + 0.01x2

C2(y) = 200 + y + y3/300

C3(z) = 200 + 10zTotal cost = C1+ C2+ C3

Find output that minimises total cost.


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Solution 1 Using substitution

z = 2000xy (constraint) total C = 200 + 0.01x2 +200 + y + y3/300

+ 200 + 10(2000xy)

= 20600 + 0.01x

2

10x + y3

/3009ypartial derivatives focCx= 0.02x10 = 0 so x = 500Cy= 0.01y

29 = 0 : y = 30 so z = 1470

soc (and total C = 17920)Cxx= 0.02 Cyy= 0.02y both >0 at (500,30,1470)

Cxy= 0 hence CxxCyy(Cxy)2> 0 min point


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Format

Constrained optimisation problem may beexpressed as an objective function andconstraint:

Minimise C = C1(x) + C2(y) + C3(z)

Objective function

s.t. x + y + z = 2000Equality constraint


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Lagrange method

1. Form = Lagrange function

2. State first order necessary conditions

x= y= z= = 0

3. Solve the above equationssimultaneously to find the critical point

4. Use second order conditions to identify

the type of critical point


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Example 2

The total revenue function for two goods x andy is given by TR = 36x3x2+ 56y4y2

and the firm is subject to a budget constraint:5x + 10y = 80

Show that y = 5.5, x = 5 are the quantities ofgoods for which first order conditions formax revenue are satisfied using theLagrange method.


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Solution 2

Problem: Max TR = 36x3x2+ 56y4y2s.t. 5x + 10y = 80

Lagrange fn:

= 36x3x2+ 56y4y2(5x + 10y80)

First order conditions:

x= 366x5 = 0 ..(i)y= 568y10= 0 ..(ii)

=5x10y + 80 = 0 .(iii)


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Solution 2 continued

From (i) = (366x)/5

From (ii) =(568y)/ 10equate the two expressions for

2(366x) = 568y

y =1.5x2substitute y =1.5x2 in (iii)

5x +15x2080 = 0

x=5 and y = 5.5 and = (36-30)/5=1.2


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Example 3

Find the values of L and K which satisfyfirst order conditions for minimum cost

when labour costs $25 per unit, capitalcosts $50 per unit when the productionconstraint is 240 units of output and theproduction function is Q = 12L0.5K0.5


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Solution 3Constrained optimisation problem is

Min Cost = 25L + 50Ks.t 12L0.5K0.5= 240

Lagrange function:

= 25L + 50K(12L0.5K0.5240)First order conditions:

L= 2512(0.5L-0.5K0.5) = 256L-0.5K0.5= 0

.(i)K= 5012(0.5L0.5K-0.5) = 506L0.5K-0.5= 0

.(ii)


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Solution 3 continued

From (i) 25 = 6 L-0.5K0.5 so =25 / 6L-0.5K0.5

From (ii) 50 = 6 L0.5K-0.5 so = 50 / 6L0.5K-0.5

Equate: 25 = 50

6L-0.5K0.5 6L0.5K-0.5

L0.5K-0.5 = 2L-0.5K0.5

L = 2K

Sub above into (iii) 240 = 12(2K)0.5K0.5

20 = 20.5K, so K = 102, L = 202

23

25

2

1

6

50

K2

K

6

50

L

K

6

50

KL6

50 5.05.05.0

5.05.0

and =


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Second order conditions - Algebraic

Identify convexity or concavity and hencewhether a min or maximum exists at the criticalpoint (x0, y0)

For f(x, y) if

fxxand fyy0 fxxfyy (fxy)2 max

fxxand fyy 0 fxxfyy (fxy)2

min

- these can be applied to the Lagrange function


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Example 4

1. Verify that the critical point from example 2is a maximum

2. Verify that the critical point from example 3is a minimum


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Solution 4(1)

x= 366x5 = 0 ..(i)

y= 568y10= 0 ..(ii)

=5x10y + 80 = 0 .(iii)

so xx= -6 yy =8 yx=0

fxx& fyy< 0 and fxxfyy-(fxy)2 0

Hence conditions for maximum are satisfied.


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Solution 4(2)

L= 256L-0.5K0.5= 0 .(i)

K= 506L0.5K-0.5= 0 .(ii)

= 24012L0.5K0.5= 0.(iii)

LL = 3L-1.5K0.5 KK=3L0.5K-1.5

KL=3L-0.5K-0.5

at K = 102; L = 202; and =(25/3)(2)

LL =325(32)-1 (202)-1.5 (102)0.5= 0.4419KK=325(32)-1 (202)0.5 (102)-1.5= 1.768

KL=325(32)-1 (202)-0.5(102)-0.5=0.8839


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Continued

LL and KK both > 0

LL KK(KL)2= 0.44191.768(-0.8836)2

> 0

Hence conditions for a minimum are satisfied

at K = 102; L = 202


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Meaning of

or = f(x,y)(g(x,y)c)

is the coefficient of c which indicates howmuch changes when c changes by 1 unit.

is the rate at which the optimal value of theobjective function changes w.r.t. an increase in

the constraint c


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Example 5 Interpreting

Given the utility function U = 5xy, find thechange in the maximum level of utility whenthe budget constraint 5x + y = 30 is increased

by one unit (to 31).


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Solution (Example 5)

Sub and 15030

When C changes by 1 unit (from 30 to 31), the

maximum level of utility changes by 15 units.

into

305(5 yxxy

yyx 055

xxy 505

0305 yx


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Recall matrix approach

Critical point (x0, y0 .) identified from firstorder conditions

2ndorder conditions for f(x, y.) |H1| < 0; |H2|>0; |H3| 0; |H2|> 0; |H3| >0.. for min

where the determinants are found at the values of critical point

How can this be extended for constrainedoptimisation ?


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Recall matrix approach

Critical point (x0, y0 .) identified from firstorder conditions

2ndorder conditions for f(x, y.) |H1| < 0; |H2|>0; |H3| 0; |H2|> 0; |H3| >0.. for min

where the determinants are found at the values of critical point

How can this be extended for constrainedoptimisation ?


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Second order conditions

Use bordered Hessian matrix, notation

Contains all the second order partial derivatives of

remember thatis equivalent to the constraint, so 2nd

order partials of

(x

& y

) are equiv to 1storder

partials of constraint g (gx& gy)

contains H and an extra row and columncontaining the first order partial derivatives of theconstraint

Format depends on how Lagrangian is set up

H

H


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For function of 2 variables:

== f(x,y) + (cg(x,y))

= = f(x,y)(g(x,y)c)

0gg

g

g

H

yx

yyyyx

xxyxx

yx

yyyyx

xxyxx

using since not

in Word equations

yyyxy

xyxxx

yx

yyyxy

xyxxx

yx

g

g

gg0

H


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2ndorder conditions using H

For max

For min

- evaluated at the critical point

0H

0H


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For Example 4 now use matrix

instead of the algebraic method1. Verify that the critical point is a maximum.

= 36x3x2+ 56y4y2(5x + 10y80)

First order conditions:

x= 366x5 = 0 (1)y= 568y10 = 0 (2)= 805x10y = 0 (3)

H

8010

065

10500

yyyxy

xyxxx

yx

g

g

gg

H

-5(-40)+10(60) > 0, hence maximum


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For Example 4 now use matrix

instead of the algebraic method

2. Determine second order conditions for minimum costusing bordered Hessian matrix.

=25L + 50K(12L0.5

K0.5

240 )

L= 256L-0.5K0.5= 0 .(i)K= 506L0.5K-0.5= 0 .(ii)

=12L0.5K0.5+ 240 = 0 .(iii)


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Solution

LL = 3L-1.5K0.5 KK=3 L

0.5K-1.5

KL=3L-0.5K-0.5

At K = 102; L= 202; =(25/3)(2),

LL = 0.4419 KK= 1.768 KL=0.8839(from last week)

L=6L-0.5K0.5 = -6 (202)-0.5 (102)0.5=4.243

k=6 (202)0.5 (102)-0.5=8.485

768.18839.0485.8

8839.04419.0243.4485.8243.40

H

H

012456.4485.815243.4

8839.04419.0485.8243.4

485.8768.18839.0485.8243.4

243.4


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Example 6

Problem: Maximise f(x,y) = ye-x

subject to ex+ ey = 6

Write the Lagrangian and first order conditions

Show that the critical point is given by x = ln 2

and y = ln 4. Use the bordered Hessian matrix to prove this

maximises the objective function.

Determine the maximum value of f(x,y). If the constraint limit changes from 6 to 7 what

effect will this have on the maximum value ofthe objective function?


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Solution (Example 6)

=y e-x(ex+ ey-6)

x= e-x- ex =0, so = e-2x

y= 1 - ey =0, so = e-y

= -ex- ey+6 =0 and y=2x

Sub in (iii) ex+ e2x-6 = 0

ex + (ex)2-6 = 0

(ex-2)(ex+3)=0ex=2 (cannot be -3) so x = ln2; y =2ln2 =ln4

= e-2x = e-2ln2=


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2ndorder derivatives:

xx= -e-xex = --.2 = -1

yy= - ey = -.eln4= -1

xy=0x= gx= e

x= 2 y=gy= ey= 4

Thus conditions for maximum are satisfied.Maximum value of objective function

= y-e-x= ln4- = 0.88629When the constraint limit changes from 6 to 7, the maximum

value of the objective function will change by , that is 0.25.

0)4(4)2(204124

1402

2104012420

H


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lect 9 calculus 4

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