lect 9 calculus 4
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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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