Constrained Maximisation IV Lagrange Function

EC115 - Methods of Economic AnalysisSpring Term, Lecture 10

Constrained Maximisation IV:Lagrange’s Function

Renshaw - Chapter 16

University of Essex - Department of Economics

Week 25

Domenico Tabasso (University of Essex - Department of Economics)Lecture 10 - Spring Term Week 25 1 / 23

Topics for this week

The Lagrange Multiplier Method: Basics

The Lagrange Multiplier Method: Application

The Lagrange Multiplier Method:Second OrderConditions


The Lagrange Multiplier Method: Basics 1

The Lagrange multiplier method’s objective is similar tothe substitution method in that it helps us convert aconstrained optimisation problem into an unconstrainedoptimisation problem.

The difference is that instead of substituting theconstraint in the objective function, we build a newfunction by adding the constraint using the so-calledLagrange multiplier.



Let’s start from our usual problem and let’s write it in themost general way:

maxx ,y

U = U(x , y)

s.t. g(x , y) = 0

We start from here and we build a completely new functionthat includes both the objective function and the constraint



This new function is called the Lagrangian and is given by:

L(x , y , λ) = U(x , y)− λg(x , y),

where λ is a completely new term called the Lagrangemultiplier. In the next slides we will treat this multiplier asa variable.



So our function L is a function of three variables: x , yand λ.

This implies that if we want to find a maximum, wehave to maximize with respect to all three variables.

Hence the first order conditions will be given by threeequations


The Lagrange Multiplier Method: Maximizing the Lagrangian Function

First Order Conditions:

∂L(x , y , λ)

∂x= 0

∂L(x , y , λ)

∂y= 0

∂L(x , y , λ)

∂λ= 0

at x = x∗, y = y ∗ and λ = λ∗.



So given the lagrangian function:

L(x , y , λ) = U(x , y)− λg(x , y),

the first order conditions are:∂L(x , y , λ)

∂x= 0 ⇒ ∂U(x , y)

∂x− λ∂g(x , y)

∂x= 0

∂L(x , y , λ)

∂y= 0 ⇒ ∂U(x , y)

∂y− λ∂g(x , y)

∂y= 0

∂L(x , y , λ)

∂λ= 0 ⇒ −g(x , y) = 0



Note that the third first order condition is simply theconstraint itself, written in the implicit way.

In order to find our optimal x∗, y ∗ and λ∗ we have to solvethe previous system of three equations in three unknowns.

This goal can be achieved making use of the fact that allthe equations are equal to 0 and hence equal to each other.


The Lagrange Multiplier Method: Application

In order to understand the properties of this maximizationmethod, let’s see an example

Consider again the consumer’s problem:

maxx ,y

U(x , y) = x1/2y 1/2

s.t. pxx + pyy = m.

The lagrangian of this problem is given by:

L(x , y , λ) = x1/2y 1/2 − λ(pxx + pyy −m).


The first order conditions are given by:

(A) :∂L(x , y , λ)

∂x= 0 ⇒ 1

2x−1/2y 1/2 − λpx = 0,

(B) :∂L(x , y , λ)

∂y= 0 ⇒ 1

2x1/2y−1/2 − λpy = 0,

(C ) :∂L(x , y , λ)

∂λ= 0 ⇒ −pxx − pyy + m = 0.


So if we rearrange equations (A) and (B) we can rewritethem as :

12x−1/2y 1/2 = λpx

12x

1/2y−1/2 = λpy

We can then simplify the two equations into a unique oneas:

12x−1/2y 1/2

12x

1/2y−1/2=λpx

λpy(1)


But:12x−1/2y 1/2

12x

1/2y−1/2=∂U/∂x∂U/∂y

and of courseλpx

λpy=

px

py

so that (1) is just the explicit way to write:

∂U/∂x∂U/∂y

=px

py(2)


Remember that (2) is a compact way to write equations Aand B from our original system which can now be rewrittenas:

∂U/∂x∂U/∂y

=px

py

m − pxx − pyy = 0

which is exactly: MRS =

px

py


Doesn’t it look familiar?!?!


Remember that (2) is a compact way to write equations Aand B from our original system which can now be rewrittenas:

∂U/∂x∂U/∂y

=px

py


which is exactly: MRS =

px

py


Doesn’t it look familiar?!?!Domenico Tabasso (University of Essex - Department of Economics)Lecture 10 - Spring Term Week 25 14 / 23

Now we can finally solve our system. So:∂U/∂x∂U/∂y

=px

py


implies y ∗

x∗=

px

py



Solving in our usual way the first equation implies that:

y ∗

x∗=

px

py=⇒ pxx∗ = pyy ∗.

In addition, (C) implies that at the solution to the firstorder conditions:

pxx∗ + pyy ∗ = m.

Combined we find that:

pxx∗ = pyy ∗ =m2

=⇒ x∗ =m2px

, y ∗ =m2py

.

which are our standard results.


How about λ?

So far we have found x∗ and y ∗. How about theoptimal value of our third variable, λ∗?

Well, once we have found x∗ and y ∗ finding λ∗ is trivial.

If we consider for example equation (A) of our initialsystem we have:

12x−1/2y 1/2 − λpx = 0


So that if we substitute in:

x∗ =m2px

and y ∗ =m2py

we obtain:

λ∗px =12

(m2px

)−1/2( m2py

)1/2

and so:λ∗ =

12

1

p1/2x p1/2

y


How about λ?

So λ∗ = 12

1p1/2x p1/2

y

Fine, but what is the economic meaning of the lagrangemultiplier, λ∗?

It measures the marginal change in the maximum utilitythe consumer can get when the constraint is weakenedor strengthened.Which means: it tells us how the maximum utilitychanges when we change m.


Here, the maximum utility the consumer can get is:

U(x∗, y ∗) =

(m2px

)1/2( m2py

)1/2

=m

2p1/2x p1/2

y

= U(px , py ,m)

where U(px , py ,m) is the indirect utility function that westudied last week.


Partially differentiating U with respect to income, m,gives:

∂U(px , py ,m)

∂m=

1

2p1/2x p1/2

y= λ∗.

Note that throughout we have assumed that we are atan interior solution: this property of λ∗ will nottypically hold when we are at a corner solution.


Second Order Conditions

It is quite messy to write down the second orderconditions for a maximum of a constrainedmaximisation (or constrained minimisation) problem.Suppose that we wish to solve:

maxx ,y

F (x , y) s.t. G (x , y) = 0.

Define the lagrangian:

L(x , y , λ) = F (x , y)− λG (x , y)

Suppose that (x = x∗, y = y ∗, λ = λ∗) solve the firstorder conditions for optimizing the lagrangian.


The second order conditions for (x = x∗, y = y ∗) to bea maximum of the original constrained maximisationproblem are then:

I ∂G∂x 6= 0;

I ∂G∂y 6= 0;

I(

∂G∂x

)2 (∂2L∂x2

)− 2

(∂G∂x

) (∂G∂y

)(∂2L∂x∂y

)+(

∂G∂y

)2 (∂2L∂y2

)≥ 0;

where all partial derivatives are evaluated at(x = x∗, y = y ∗, λ = λ∗).

Evidently very messy: we can happily forget aboutthem!


The second order conditions for (x = x∗, y = y ∗) to bea maximum of the original constrained maximisationproblem are then:

I ∂G∂x 6= 0;

I ∂G∂y 6= 0;

I(

∂G∂x

)2 (∂2L∂x2

)− 2

(∂G∂x

) (∂G∂y

)(∂2L∂x∂y

)+(

∂G∂y

)2 (∂2L∂y2

)≥ 0;

where all partial derivatives are evaluated at(x = x∗, y = y ∗, λ = λ∗).Evidently very messy: we can happily forget aboutthem!


Constrained Maximisation IV Lagrange Function

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