Constrained Maximisation IV- Examples

EC115 - Methods of Economic AnalysisSpring Term, Lecture 10

Constrained Maximisation IV:Lagrange’s Function - Examples

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 / 26

Example 1

Let’s start with a very simple example, just to warm up.A firm has to maximize the output under the total costconstraint TC=100. The maximization problem is given by

maxK ,L

Q(K , L) = 2K + KL

s.t. 100 = 2K + 2L

where the cost of capital and and that of labour are givenby w = r = 2.Find the optimal K ∗ and L∗ using the Lagrangian method.Find also λ∗.


First Step: Write down the Lagrangian function

L = Q(K , L)− λ(TC (K , L))

And so in this case:

L = 2K + KL− λ(2K + 2L− 100)


Second Step: Write down the First Order Conditions

∂L∂K

= 0 ⇒ 2 + L− 2λ = 0 (1)

∂L∂L

= 0 ⇒ K − 2λ = 0; (2)

∂L∂λ

= 0 ⇒ 100− 2K − 2L = 0 (3)


Third Step: Solve the First Order Conditions.From equations (1) and (2) we have

2 + L = 2λ (1)

K = 2λ (2)

The right-hand sides of both equations are identical (= 2λ),so the left-hand must also be equal to each other. Hence

2 + L = K


Now our problem only consists in two equations:{2 + L = K2K + 2L = 100

So substituting K = 2 + L form the first equation into thesecond one we get:

4 + 2L + 2L = 100=⇒

L∗ = 24


So L∗ = 24 and we know that K = 2 + L: we can obtainK ∗ = 26.

How about λ?Form the second first order condition we have:

2λ = K

⇒ λ =K2

⇒ λ∗ = 13


Example 2

Note that sometimes very simple changes can make oursolution very complicated! Consider for example

maxK ,L

Q(K , L) = 2K + 2K12L

s.t. 100 = 2K + 2L

and proceed to the maximization following exactly thesame steps as before.


Write down the Lagrangian function

L = 2K + 2K12L− λ(2K + 2L− 100)

and the FOCS:

∂L∂K

= 0 ⇒ 2 + K−12L− 2λ = 0 (1)

∂L∂L

= 0 ⇒ 2K12 − 2λ = 0; (2)

∂L∂λ

= 0 ⇒ 100− 2K − 2L = 0 (3)


Solve the First Order Conditions.In this case simply taking the ratio between eq. (1) andeq.(2) is not very helpful since if we do it we end up with:

2 + K−12L

2K12

=2λ2λ

where the left-hand side can not be simplified. Let’s see ifwe can find another way to solve these equations.


From equation (1) and (2) we have

2 + K−12L = 2λ (1)

2K12 = 2λ (2)

The right-hand sides of both equations are identical (= 2λ),so the left-hand must also be equal to each other. Hence

2 + K−12L = 2K

12


Multiplying everything by K12 we can get:

2K12 + L = 2K

14

from which we can obtain that

L∗ = 2K14 − 2K

12 (4)

Don’t forget we have another equation to play with, thethird first order condition.

−2K − 2L + 100 = 0


Solving the constraint for L we get:

L = 50− K

which we can substitute in eq. (4), obtaining:

2K14 − 2K

12 = 50− K


OK, now we have one (complicated, sorry /) equation inone unknown. We can try to solve it.

50− K − 2K14 + 2K

12 = 0

K + 2K14 − 2K

12 = 50

Scary!


Example: Cost Minimisation

Suppose the board of directors of a competitive firmhas set a production target of Q0 and asks the chiefexecutive officer (CEO) of the firm to meet the targetat the minimum cost.Suppose the firm’s technology be described by theCobb Douglas function:

Q = F (K , L) = K 1/4L1/2

How do we solve the CEO problem when the prices oflabour and capital are given by w > 0 and r > 0?

We can apply any of our solution techinques to solvethis problem. This time we use the Lagrange multipliermethod.


The problem can be expressed as:

minL,K

TC (K , L) = wL + rK s.t. Q0 = K 1/4L1/2

In this case the objective function is given by:

TC = wL + rK .

This describes all the combinations of K and L suchthat at prices w and r they give the same total cost,TC .In this case we say that TC = wL + rK describes aniso-cost line:

K = −wr

L +TCr.


The lagrangian function is given by:

L(L,K , λ) = wL + rK + λ(Q0 − K 1/4L1/2)

The first order conditions are:

∂L(K , L, λ)

∂K= r − λ1

4K−3/4L1/2 = 0,

∂L(K , L, λ)

∂L= w − λ1

2K 1/4L−1/2 = 0,

∂L(K , L, λ)

∂λ= Q0 − K 1/4L1/2 = 0,

at K = K ∗, L = L∗ and λ = λ∗.


Note that the first two equations generate the tangencycondition:

MRTSL,K =MPL

MPK=

wr

=⇒(12K 1/4L−1/2

)/

(14K−3/4L1/2

)=

wr.

The third equation requires that the targeted outputlevel met:

Q0 = (K ∗)1/4 (L∗)1/2 .


From the tangency condtion:

2K ∗

L∗=

wr

=⇒ K ∗ = L∗(w2r

)Using the constraint:

Q0 =(L∗

(w2r

))1/4(L∗)1/2 = (L∗)3/4

(w2r

)1/4

Solving for L∗ and K ∗ we obtain:

L∗ = Q4/30

(2rw

)1/3

,

K ∗ = L∗(w2r

)=⇒ K ∗ = Q4/3

0

(w2r

)2/3


Substituting into the expression for total costs we getthe minimised total cost:

TC ∗ = wL∗ + rK ∗ =3

22/3Q4/30 w 2/3r 1/3.

Now consider the effect of increasing the wage rate:

∂L∗

∂w= −1

321/3Q4/3

0 w−4/3r 1/3 < 0,

∂K ∗

∂w=

232−2/3Q4/3

0 w−1/3r−2/3 > 0.

If we increase w from w1 to w2 then L∗ goes down, K ∗

goes up: we substitute capital for labour.


Also note that from the first order conditions:

λ∗ =4r(K ∗)3/4

(L∗)1/2 =4r [Q4/3

0 (w/(2)r)2/3]3/4

[Q4/30 (2r/w)1/3]1/2

= 24/3Q1/30 w 2/3r 1/3.

This is the same as the marginal cost of production atQ0 using cost minimising input choices, i.e. ∂TC∗

∂Q0.


Example: Profit Maximisation

Now suppose the board of directors of a competitivefirm wants to maximise profits and asks the CEO of thefirm to perform this task by choosing the appropriatelevels of labour and capital.In this case the CEO’s constraint is the firm’stechnology. Suppose this is described by the CobbDouglas function:

Q = F (K , L) = K 1/4L1/2

How to solve the CEO problem when the price of thegood the firm sells is p and the prices of labour andcapital are given by w > 0 and r > 0?


We could use an unconstrained maximisation approachin which we we want to maximise:

Π(Q,K , L) = pQ − wL− rK

and we substitute out the constraint and then solve theproblem:

maxK ,L

Π(K , L) = pK 1/4L1/2 − wL− rK .


Alternatively we could use a constrained maximisationapproach and maximise profit (equals revenue minuscost) subject to output equaling sales:

maxQ,K ,L

pQ − wL− rK s.t. Q = F (K , L).

Using the Lagrange Multiplier Method then we solvethe following problem:

maxQ,K ,L,λ

L(Q,K , L, λ) = pQ−wL−rK +λ(Q−K 1/4L1/2)


The first order conditions are then:

A :∂L∂Q

= p − λ = 0,

B :∂L∂K

= −r + λ

(14

)K−3/4L1/2 = 0,

C :∂L∂L

= −w + λ

(12

)K 1/4L−1/2 = 0,

D :∂L∂λ

= Q − K 1/4L1/2 = 0.

Denoting the solution as(Q = Q∗,K = K ∗, L = L∗, λ = λ∗) then from A we getλ∗ = p.


Substituting into B and C gives:

p(14

)K−3/4L1/2 = r , p

(12

)K 1/4L−1/2 = w .

Then:

L1/2 =4rK 3/4

p=

pK 1/4

2w,

so:

K 1/2 =p2

8rw=⇒ K ∗ =

p4

64r 2w 2

and:

L∗ =p2K 1/2

4w 2 =p4

32rw 3


Constrained Maximisation IV- Examples

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