Optimization With More Than One Variable

EC115 - Methods of Economic AnalysisLecture 4

Optimization with more than one variable

Renshaw - Chapter 15

University of Essex - Department of Economics

Week 19

Domenico Tabasso (University of Essex - Department of Economics)Lecture 4 Week 19 1 / 37

Introduction

A function of several variables is a relation betweensome independent variables x1, x2, x3, ...xn and somedependent variable z such that:

z = f (x1, x2, x3, ...xn)

specifies the value of z given the values ofx1, x2, x3, ...xn.Today we will analyze the maximum and minimumvalues of these type of functions.We do this by building on our knowledge of partialderivatives.


Stationary Points

Consider the function z = f (x , y). The point(x = x0, y = y0) is called a stationary point of z if:

∂z∂x

=∂z∂y

= 0 at (x = x0, y = y0).

This is just like the definition of a stationary point for afunction of one variable: there the function w = g(x)had a stationary point at x = x0 if dw

dx = 0 at x = x0.However, when we deal with functions of more than onevariable then it turns that there are three kinds ofstationary points: maximum points, minimum pointsand saddlepoints.


Conditions for a Minimum

Consider the following function

z = (x2 + y 2).

The first and second order partial derivatives are givenby

∂z∂x

= 2x ;∂2z∂x2 = 2,

∂z∂y

= 2y ;∂2z∂y 2 = 2,

∂2z∂x∂y

=∂2z

∂y∂x= 0.


Can we use these derivatives to establish whether thefunction has a minimum or a maximum?


Conditions for a Minimum

The function clearly has a minimum. In order to study itwe first assume that one of the variables is a constant.

Assume y = y0 = 0. Graph this function. What is thevalue of x that makes z as small as possible?

I Here f (x , y0) = f (x , 0) = x2 which is minimized with respect to xwhen x = 0.

Assume x = x0 = 0. Graph this function. What is thevalue of y that makes z as small as possible?

I Here f (x0, y) = f (0, y) = y2 which is minimized with respect to ywhen y = 0.


Using the iso-x section to find necessary conditions for a minimum


Note that for any value of y , the function decreases forall values of x < 0 and then increases.

∂z∂x

< 0 for all x < 0 and∂z∂x

> 0 for all x > 0.

Also note that for any value of x , the functiondecreases for all values of y < 0 and then increases.

∂z∂y

< 0 for all y < 0 and∂z∂y

> 0 for all y > 0.

Note that in both cases the second order partialderivative is always positive.


The first order conditions (FOC) for z = f (x , y) tohave a minimum at (x = x0, y = y0) are given by:

∂z∂x

= 0 at (x = x0, y = y0),

∂z∂y

= 0 at (x = x0, y = y0).


The second order conditions (SOC) for z = f (x , y) tohave a minimum at (x = x0, y = y0) are given by:

∂2z∂x2 > 0 at (x = x0, y = y0),

∂2z∂y 2 > 0 at (x = x0, y = y0),

∂2z∂x2 ×

∂2z∂y 2 >

∂2z∂x∂y

× ∂2z∂y∂x

=

(∂2z

∂x∂y

)2

at (x = x0, y = y0),

where the last equality follows from Young’s Theorem.


The first order conditions and the first two parts of thesecond order conditions for z = f (x , y) to have aminimum at (x = x0, y = y0) are the same as the firstand second order conditions for f (x , y0) to have aminimum with respect to x at x = x0 and f (x0, y) tohave a minimum with respect to y at y = y0.

The third part of the second order conditions is neededin addition to the first two parts because we can vary xand y in other ways than just keeping one of them fixedwhile changing the other.


Conditions for a Maximum

Consider the following function:

z = −(x2 + y 2).

The first and second order partial derivatives are givenby:

∂z∂x

= −2x ;∂2z∂x2 = −2

∂z∂y

= −2y ;∂2z∂y 2 = −2

∂2z∂x∂y

=∂2z

∂y∂x= 0.



Assume y = y0 = 0. Graph this function. What is thevalue of x that makes z as big as possible?

I Here f (x , y0) = f (x , 0) = −x2 which is maximized with respect to xwhen x = 0.

Assume x = x0 = 0. Graph this function. What is thevalue of y that makes z as big as possible?

I Here f (x0, y) = f (0, y) = −y2 which is maximized with respect to ywhen y = 0.


For any value of y , the function is increasing for allvalues of x < 0 and then decreases.

∂z∂x

> 0 for all x < 0 and∂z∂x

< 0 for all x > 0.

For any value of x , the function is increasing for allvalues of y < 0 and then decreases.

∂z∂y

> 0 for all y < 0 and∂z∂y

< 0 for all y > 0.

Note that in both cases the second order own partialderivative is always negative.


Using the iso-y section to find necessary conditions for a maximum


The first order conditions for z = f (x , y) to have amaximum at (x = x0, y = y0) are given by:

∂z∂x

= 0 at (x = x0, y = y0),

∂z∂y

= 0 at (x = x0, y = y0).


The second order conditions for z = f (x , y) to have amaximum at (x = x0, y = y0) are given by:

∂2z∂x2 < 0 at (x = x0, y = y0),

∂2z∂y 2 < 0 at (x = x0, y = y0),

∂2z∂x2

∂2z∂y 2 >

∂2z∂x∂y

∂2z∂y∂x

=

(∂2z

∂x∂y

)2

at (x = x0, y = y0),

where, again, the last equality follows from Young’sTheorem.


Example of Failure of SOC for a Maximum

Consider the following function:

z = (x + y)2 − 2(x − y)2 = −x2 + 6xy − y 2.

The first and second order partial derivatives are givenby:

∂z∂x

= 6y − 2x ;∂2z∂x2 = −2;

∂z∂y

= 6x − 2y ;∂2z∂y 2 = −2;

∂2z∂x∂y

=∂2z

∂y∂x= 6.



At x = y = 0 then ∂z∂x = 0, ∂2z

∂x2 = −2, ∂z∂y = 0,

∂2z∂y2 = −2.

Therefore the value of x which makes z as large aspossible when y = 0 is x = 0 and, similarly, the valueof y which makes z as large as possible when x = 0 isy = 0. In addition, x = y = 0 is the only solution tothe first order conditions.



However, since ∂2z∂x∂y = ∂2z

∂y∂x = 6 then(

∂2z∂x∂y

)2= 36

while ∂2z∂x2 × ∂2z

∂y2 = 4 < 36 so that the third part of thesecond order conditions is not satisfied.

At x = y = 0 then z = 0 but at x = y = a thenz = 4a2 which is greater than zero whenever a 6= 0.Therefore, z does not have a maximum at x = y = 0.


Need for third part of 2nd order conditions


Conditions for a Saddle point

The first order conditions for z = f (x , y) to have asaddlepoint at (x = x0, y = y0) are given by:

∂z∂x

= 0 at (x = x0, y = y0),

∂z∂y

= 0 at (x = x0, y = y0).


The second order conditions for z = f (x , y) to have asaddlepoint at (x = x0, y = y0) are given by:

∂2z∂x2

∂2z∂y 2 <

∂2z∂x∂y

∂2z∂y∂x

=

(∂2z

∂x∂y

)2

at (x = x0, y = y0),

where, again, the last equality follows from Young’sTheorem.

This condition is automatically met if ∂2z∂x2 and ∂2z

∂y2 haveopposite signs.


Example of a saddle point


Strategy for Optimization

Identify locations of stationary points by determiningwhere first order conditions are satisfied.

I The first order conditions are necessary conditions for a stationarypoint.

For each stationary point determine whether it is of thedesired type (maximum, minimum, saddlepoint) byexamining whether or not it satisfies the appropriatesecond-order conditions.

I The appropriate second-order conditions combined with the first orderconditions are sufficient conditions for the desired type of stationarypoint.


A few examples - 1Find the maximum or the minimum of the followingfunction:

z = f (x , y) = 6− x2 − 2xy − 3y 2 − 3x + 4y ;

First Step: Search for all the critical points of thefunction =⇒ look for all the points that satisfy the firstorder conditions

FOCs∂z∂x

= 0⇒ −2x − 2y − 3 = 0

∂z∂y

= 0⇒ −2x − 6y + 4 = 0


Example 1 - Cont.

The FOCs are represented by two equations, simultaneouslyequal to 0. We solve the as a system and the solutions aregoing to represent the candidate points for maxima and/orminima.

Second Step: Solve the FOCs{−2x − 2y − 3 = 0−2x − 6y + 4 = 0

⇒{

x = −32 − y

3 + 2y − 6y + 4 = 0

⇒{

x∗ = −134

y ∗ = 74


Example 1 - Cont.

So the point(−13

4 , 74

)is the only candidate. Is it a max or

a min (or maybe a saddle point)?

Third Step: Check the signs of ALL the secondorder conditions:

∂2z∂x2 > < 0 ?

∂2z∂y 2 > < 0 ?

∂2z∂x2 ×

∂2z∂y 2 > <

(∂2z

∂x∂y

)2

?


Example 1 - Cont.In our case:

∂2z∂x2 = −2 < 0 ∀ x , y

∂2z∂y 2 = −6 < 0 ∀ x , y

It looks like we’re dealing with a maximum. Really? YES:

(−2)×(−6) =∂2z∂x2×

∂2z∂y 2 = 12 >

(∂2z

∂x∂y

)2

= (−2)2 = 4

So, the function:z = f (x , y) = 6− x2 − 2xy − 3y 2 − 3x + 4y has amaximum at

(−13

4 , 74

)Domenico Tabasso (University of Essex - Department of Economics)Lecture 4 Week 19 30 / 37

Economic Applications

A firm produces good Q in a competitive market and sellsit at price P . This firm need both capital and labour to beable to produce any quantity of Q. It hires labour at awage of w and rents capital at a rate of r . Let theproduction function of the firm be Cobb-Douglas:

Q = K 1/2L1/3.

Write down the firm’s profit function.What are the demands for labour and capital if thisfirm maximizes profits?Graph your answer.


Solution - 1

The profit function is

Π = p ∗ K 1/2L1/3 − wL− rK

and the first order conditions associated with it are:

∂Π

∂K= 0⇒ 1

2pK−1/2L1/3 − r = 0

∂Π

∂L= 0⇒ 1

3pK 1/2L−2/3 − w = 0


Solution - 2

Solving simultaneously the two FOCs we can obtain:

12pK−1/2L1/3

13pK 1/2L−2/3

=rw

By simplifying and solving with respect to L we get:

L =23

rw

K (1)

which we can now plug back into any of the two FOCs.


Solution - 3

Substituting (1) into the first FOC we obtain:

12pK−1/2

(23

rw

K)1/3

− r = 0

that can be solved for K in order to get

K ∗ =p6

144r 4w 2

Substituting K∗ into (1) we eventually obtain:

L∗ =p6

216r 3w 3


Solution - 4

So the point(K ∗ = p6

144r4w2 , L∗ = p6

216r3w3

)is a critical

point. Is it a max? Let’s check the second order conditions.

∂2Π∂K 2 = −1

4pK−3/2L1/3 < 0, for any positive p, K , L

∂2Π∂L2 = −2

9pK−1/2L−5/3 < 0, for any positive p, K , L

So the first two SOCs are satisfied, for any value of K andL, and hence also for K ∗, L∗ (remember that K and L areproduction inputs, so assuming that they are alwayspositive is a very mild hypothesis).


Solution - 5

Furthermore:

∂2Π∂K∂L = 1

6pK−1/2L−2/3 > 0, for any positive p, K , L

so:

∂2Π

∂K 2 ×∂2Π

∂L2 =

(−14pK−3/2L1/3

)×

(−29pK−1/2L−5/3

)=

236

p2K−1L−4/3

>

136

p2K−1L−4/3 =

(∂2Π

∂K∂L

)2


Solution - 6

So, as all the necessary second order conditions aresatisfied, the point

(K ∗ = p6

144r4w2 , L∗ = p6

216r3w3

)indeed

represents a maximum.


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Optimization With More Than One Variable