2 THE MATHEMATICS OF OPTIMIZATIONBBM

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Chapter 2

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The Mathematics of Optimization

• Many economic theories begin with the assumption that an economic agent is

seeking to find the optimal value of some function

– consumers seek to maximize utility – firms seek to maximize profit

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Maximization of a Function of

One Variable

• Simple example: Manager of a firm wishes to maximize profits

) (q f

 

 = f(q)



*

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Maximization of a Function of

One Variable

• The manager will likely try to vary q to see where the maximum profit occurs

– an increase from q₁ to q₂ leads to a rise in _

 = f(q)



*

1

2



0



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Maximization of a Function of

One Variable

• If output is increased beyond q*, profit will decline

– an increase from q* to q₃ leads to a drop in _

 = f(q)



*

0



q

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Derivatives

• The derivative of _ = f(q) is the limit of /q for very small changes in q

h q f h q f dq df dq d h ) ( ) (

lim 1 1

0      

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Value of a Derivative at a Point

• The evaluation of the derivative at the point q = q1 can be denoted

1 q q dq d  

• In our previous example,

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First Order Condition for a

Maximum

• For a function of one variable to attain its maximum value at some point, the derivative at that point must be zero

0 

q*

q

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Second Order Conditions

• The first order condition (d_/dq) is a

necessary condition for a maximum, but it is not a sufficient condition



*

If the profit function was u-shaped, the first order condition would result in q* being chosen and _ would

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Second Order Conditions

• This must mean that, in order for q* to be the optimum,

*

q q

dq d

 

 ₀ _for

and _dqd  0 for q  q *

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Second Derivatives

• The derivative of a derivative is called a second derivative

• The second derivative can be denoted by

) ( " or or ₂

2 2

2

q f

dq f d dq

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Second Order Condition

• The second order condition to represent a (local) maximum is

0 )

(

" _*

* 2

2

 



 

q q q

q

q f

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Rules for Finding Derivatives

0 then constant, a is If 1. _ dx db b ) ( ' )] ( [ then constant, a is If

2. bf x

dx x bf d b _ 1 then constant, is If 3.   b b bx dx dx b x

d ln 1

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a a

a dx

dax _x

constant any

for ln

5. _

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Rules for Finding Derivatives

) ( ' ) ( ' )] ( ) ( [

6. f x g x

dx x g x f d    ) ( ) ( ' ) ( ' ) ( )] ( ) ( [

7. f x g x f x g x dx x g x f d   

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dz dg dx

df dz

dx dx

dy dz

dy

 

9.

• If y = f(x) and x = g(z) and if both f’(x) and g’(x) exist, then:

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ax ax ax ax ae a e dx ax d ax d de dx de      ( ) ) ( 10.

• Some examples of the chain rule include

) ln( ) ln( ) ( ) ( )] [ln( )] [ln(

11. ax a a ax

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Example of Profit Maximization

• Suppose that the relationship between profit and output is

 = 1,000q - 5q2

• The first order condition for a maximum is

d_/dq = 1,000 - 10q = 0

q* = 100

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Functions of Several Variables

• Most goals of economic agents depend on several variables

– trade-offs must be made

• The dependence of one variable (y) on a series of other variables (x1,x2,…,xn) is

denoted by

) ,...,

,

(x x x_n f

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• The partial derivative of y with respect to

x1 is denoted by

Partial Derivatives

1 1

1

1 f

f x

y

x or or

or 

 



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• A more formal definition of the partial derivative is

Partial Derivatives

h x x x f x x h x f x

f n n

h x x n ) ,..., , ( ) ,..., , (

lim 1 2 1 2

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Calculating Partial Derivatives

2 1 2 2 2 1 1 1 2 2 2 1 2 1 2 1 2 2 cx bx f x f bx ax f x f cx x bx ax x x f y               and then , ) , ( If 1. 2 1 2 1 2 1 2 1 2 1 bx ax bx ax bx ax be f x f ae f x f e x x f y            and then If

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Calculating Partial Derivatives

2 2 2 1 1 1 2 1 2 1 x b f x f x a f x f x b x a x x f y            and then If

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Partial Derivatives

• Partial derivatives are the mathematical expression of the ceteris paribus

assumption

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Partial Derivatives

• We must be concerned with how variables are measured

– if q represents the quantity of gasoline

demanded (measured in billions of gallons) and p represents the price in dollars per

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Elasticity

• Elasticities measure the proportional effect of a change in one variable on another

– unit free

• The elasticity of y with respect to x is

y x x y y x x y x y y

e_y _x _

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Elasticity and Functional Form

• Suppose that

y = a + bx + other terms

• In this case,

             bx a x b y x b y x x y e_y_,_x

• ey,x is not constant

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Elasticity and Functional Form

• Suppose that

y = axb • In this case,

b ax

x abx

y x x

y

e_y _x _ _ b _ _b _

 

 1

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Elasticity and Functional Form

• Suppose that

ln y = ln a + b ln x

• In this case,

x y b y x x y e_y _x

ln ln ,        

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Second-Order Partial Derivatives

• The partial derivative of a partial derivative is called a second-order partial derivative

ij i

j j

i _f

x x

f x

x f

 

  

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Young’s Theorem

• Under general conditions, the order in which partial differentiation is conducted to evaluate second-order partial

derivatives does not matter

ji ij

f

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Use of Second-Order Partials

• Second-order partials play an important role in many economic theories

• One of the most important is a variable’s own second-order partial, f_ii

– shows how the marginal influence of xi on

y(_y/_xi) changes as the value of xi

increases

– a value of fii < 0 indicates diminishing

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Total Differential

• Suppose that y = f(x₁,x₂,…,x_n)

• If all x’s are varied by a small amount, the total effect on y will be

n n dx x f dx x f dx x f dy         

 2 ...

2 1 1 n ndx f dx f dx f

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First-Order Condition for a

Maximum (or Minimum)

• A necessary condition for a maximum (or minimum) of the function f(x1,x2,…,xn) is

that dy = 0 for any combination of small changes in the x’s

• The only way for this to be true is if

0 ...

2

1  f   fn 

f

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Finding a Maximum

• Suppose that y is a function of x1 and x2

y = - (x1 - 1)2 - (x2 - 2)2 + 10

y = - x12 + 2x1 - x22 + 4x2 + 5

• First-order conditions imply that

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Production Possibility Frontier

• Earlier example: 2x2 + y2 = 225

• Can be rewritten: f(x,y) = 2x2 + y2 - 225 = 0

• Because fx = 4x and fy = 2y, the opportunity

cost trade-off between x and y is

y x y

x f

f dx

dy

y

x 2

2

4 _

 

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Implicit Function Theorem

• It may not always be possible to solve implicit functions of the form g(x,y)=0 for unique explicit functions of the form y = f(x)

– mathematicians have derived the necessary conditions

– in many economic applications, these

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The Envelope Theorem

• The envelope theorem concerns how the optimal value for a particular function

changes when a parameter of the function changes

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The Envelope Theorem

• Suppose that y is a function of x

y = -x2 + ax

• For different values of a, this function

represents a family of inverted parabolas • If a is assigned a specific value, then y

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The Envelope Theorem

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As a increases,

the maximal value for y (y*) increases

The relationship between a and y

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The Envelope Theorem

• Suppose we are interested in how y* changes as a changes

• There are two ways we can do this

– calculate the slope of y directly

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• To calculate the slope of the function, we must solve for the optimal value of x for any value of a

dy/dx = -2x + a = 0

x* = a/2

• Substituting, we get

y* = -(x*)2 + a(x*) = -(a/2)2 + a(a/2)

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• Therefore,

dy*/da = 2a/4 = a/2 = x*

• But, we can save time by using the envelope theorem

– for small changes in a, dy*/da can be

computed by holding x at x* and calculating

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The Envelope Theorem

y/ a = x

• Holding x = x*

y/ a = x* = a/2

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• The envelope theorem states that the

change in the optimal value of a function with respect to a parameter of that function can be found by partially differentiating the objective function while holding x (or

several x’s) at its optimal value

)}

(

*

{

*

a

x

a

y

da

dy



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• The envelope theorem can be extended to the case where y is a function of several variables

y = f(x1,…xn,a)

• Finding an optimal value for y would consist of solving n first-order equations

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The Envelope Theorem

• Optimal values for theses x’s would be determined that are a function of a

x1* = x1*(a)

x2* = x2*(a)

x_n*= x_n*(a)

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• Substituting into the original objective

function yields an expression for the optimal value of y (y*)

y* = f [x1*(a), x2*(a),…,xn*(a),a]

• Differentiating yields

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• Because of first-order conditions, all terms except _f/_a are equal to zero if the x’s are at their optimal values

• Therefore,

)} (

* {

*

a x

x a

f da

dy

 

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Constrained Maximization

• What if all values for the x’s are not feasible?

– the values of x may all have to be positive – a consumer’s choices are limited by the

amount of purchasing power available

• One method used to solve constrained

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Lagrangian Multiplier Method

• Suppose that we wish to find the values of x1, x2,…, xn that maximize

y = f(x₁, x₂,…, x_n)

subject to a constraint that permits only certain values of the x’s to be used

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Lagrangian Multiplier Method

• The Lagrangian multiplier method starts with setting up the expression

L = f(x1, x2,…, xn ) + g(x1, x2,…, xn)

where _ is an additional variable called a Lagrangian multiplier

• When the constraint holds, L = f

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• First-Order Conditions

L/x1 = f1 + g1 = 0

L/x2 = f2 + g2 = 0

.

L/x_n = f_n + g_n = 0 .

.

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• The first-order conditions can generally be solved for x1, x2,…, xn and 

• The solution will have two properties:

– the x’s will obey the constraint

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Lagrangian Multiplier Method

• The Lagrangian multiplier (_) has an important economic interpretation

• The first-order conditions imply that

f1/-g1 = f2/-g2 =…= fn/-gn = 

– the numerators above measure the marginal benefit that one more unit of xi will have for the function f

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• At the optimal choices for the x’s, the

ratio of the marginal benefit of increasing

x_i to the marginal cost of increasing x_i

should be the same for every x

  is the common cost-benefit ratio for all of the x’s

i i

x x

of cost marginal

of benefit marginal

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Lagrangian Multiplier Method

• If the constraint was relaxed slightly, it would not matter which x is changed • The Lagrangian multiplier provides a

measure of how the relaxation in the constraint will affect the value of y

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• A high value of _ indicates that y could be increased substantially by relaxing the constraint

– each x has a high cost-benefit ratio

• A low value of _ indicates that there is not much to be gained by relaxing the constraint

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Duality

• Any constrained maximization problem has associated with it a dual problem in constrained minimization that focuses attention on the constraints in the

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Duality

• Individuals maximize utility subject to a budget constraint

– dual problem: individuals minimize the

expenditure needed to achieve a given level of utility

• Firms minimize the cost of inputs to produce a given level of output

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Constrained Maximization

• Suppose a farmer had a certain length of

fence (P) and wished to enclose the largest

possible rectangular shape

• Let x be the length of one side

• Let y be the length of the other side

• Problem: choose x and y so as to maximize

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Constrained Maximization

• Setting up the Lagrangian multiplier

L = x·y + _(P - 2x - 2y)

• The first-order conditions for a maximum are

L/x = y - 2 = 0

L/y = x - 2 = 0

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Constrained Maximization

• Since y/2 = x/2 = _, x must be equal to y

– the field should be square

– x and y should be chosen so that the ratio of marginal benefits to marginal costs should be the same

• Since x = y and y = 2_, we can use the constraint to show that

x = y = P/4

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• Interpretation of the Lagrangian multiplier

– if the farmer was interested in knowing how

much more field could be fenced by adding an extra yard of fence, _ suggests that he could find out by dividing the present perimeter (P) by 8

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Constrained Maximization

• Dual problem: choose x and y to minimize the amount of fence required to surround the field

minimize P = 2x + 2y subject to A = x·y

• Setting up the Lagrangian:

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• First-order conditions:

LD/_x = 2 - _D·y = 0

LD/_y = 2 - _D·x = 0

LD/_D = A - x·y = 0

• Solving, we get

x = y = A1/2

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Envelope Theorem &

• Suppose that we want to maximize y = f(x1,…,xn;a)

subject to the constraint

g(x1,…,xn;a) = 0

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Envelope Theorem &

• Alternatively, it can be shown that

dy*/da = _L/_a(x1*,…,xn*;a)

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Inequality Constraints

• In some economic problems the constraints need not hold exactly • For example, suppose we seek to

maximize y = f(x₁,x₂) subject to g(x1,x2)  0,

x₁ _ 0, and

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• One way to solve this problem is to

introduce three new variables (a, b, and

c) that convert the inequalities into equalities

• To ensure that the inequalities continue to hold, we will square these new

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Inequality Constraints

g(x1,x2) - a2 = 0;

x1 - b2 = 0; and

x2 - c2 = 0

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• We can set up the Lagrangian

L = f(x₁,x₂) + _₁[g(x₁,x₂) - a2] + _₂[x₁- b2] + _₃[x₂

- c2]

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L/x1 = f1 + 1g1 + 2 = 0

L/x2 = f1 + 1g2 + 3 = 0

L/a = -2a1 = 0

L/b = -2b2 = 0

L/c = -2c3 = 0

L/1 = g(x1,x2) - a2 = 0

L/2 = x1 - b2 = 0

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• According to the third condition, either a

or _1 = 0

– if a = 0, the constraint g(x1,x2) holds exactly

– if _₁ = 0, the availability of some slackness of the constraint implies that its value to the objective function is 0

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Inequality Constraints

• These results are sometimes called Kuhn-Tucker conditions

– they show that solutions to optimization problems involving inequality constraints will differ from similar problems involving equality constraints in rather simple ways – we cannot go wrong by working primarily

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Second Order Conditions -

Functions of One Variable

• Let y = f(x)

• A necessary condition for a maximum is that

dy/dx = f ’(x) = 0

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Second Order Conditions -

• The total differential measures the change in y

dy = f ’(x) dx

• To be at a maximum, dy must be decreasing for small increases in x

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• Note that d 2y < 0 implies that f ’’(x)dx2 < 0

• Since dx2 must be positive, f ’’(x) < 0

• This means that the function f must have a concave shape at the critical point

2 2 [ '( ) ] _dx _f _"₍_x₎_dx _dx _f _"₍_x₎_dx

dx

dx x

f d y

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Second Order Conditions -

Functions of Two Variables

• Suppose that y = f(x1, x2)

• First order conditions for a maximum are y/x1 = f1 = 0

y/x2 = f2 = 0

• To ensure that the point is a maximum, y

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• The slope in the x₁ direction (f₁) must be diminishing at the critical point

• The slope in the x₂ direction (f₂) must be diminishing at the critical point

• But, conditions must also be placed on the cross-partial derivative (f12 = f21) to ensure

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Second Order Conditions -

• The total differential of y is given by

dy = f₁ dx₁ + f₂ dx₂

• The differential of that function is

d 2y = (f₁₁dx₁ + f₁₂dx₂)dx₁ + (f₂₁dx₁ + f₂₂dx₂)dx₂

d 2y = f₁₁dx₁₂ + f₁₂dx₂dx₁ + f₂₁dx₁dx₂ + f₂₂dx₂₂

• By Young’s theorem, f₁₂ = f₂₁ and

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d 2y = f₁₁dx₁₂ + 2f₁₂dx₁dx₂ + f₂₂dx₂₂

• For this equation to be unambiguously

negative for any change in the x’s, f11 and f22

must be negative

• If dx₂ = 0, then d 2y = f₁₁ dx₁₂

– for d 2y < 0, f₁₁ < 0

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Second Order Conditions -

d 2y = f₁₁dx₁₂ + 2f₁₂dx₁dx₂ + f₂₂dx₂₂

• If neither dx1 nor dx2 is zero, then d 2y will be

unambiguously negative only if

f11 f22 - f122 > 0

– the second partial derivatives (f11 and f22) must

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cross-Constrained Maximization

• Suppose we want to choose x₁ and x₂ to maximize

y = f(x1, x2)

• subject to the linear constraint c - b1x1 - b2x2 = 0

• We can set up the Lagrangian

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Constrained Maximization

• The first-order conditions are f1 - b1 = 0

f2 - b2 = 0

c - b1x1 - b2x2 = 0

• To ensure we have a maximum, we must use the “second” total differential

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• Only the values of x1 and x2 that satisfy the constraint can be considered valid alternatives to the critical point

• Thus, we must calculate the total differential of the constraint

-b1 dx1 - b2 dx2 = 0

dx₂ = -(b₁/b₂)dx₁

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Constrained Maximization

• Because the first-order conditions imply that f₁/f₂ = b₁/b₂, we can substitute and get

dx₂ = -(f₁/f₂) dx₁

• Since

d 2y = f₁₁dx₁₂ + 2f₁₂dx₁dx₂ + f₂₂dx₂₂

we can substitute for dx₂ and get

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• Combining terms and rearranging

d 2y = f₁₁f₂₂- 2f₁₂f₁f₂ + f₂₂f₁₂[dx₁₂/ f₂₂]

• Therefore, for d 2y < 0, it must be true that

f₁₁f₂₂- 2f₁₂f₁f₂ + f₂₂f₁₂ < 0

• This equation characterizes a set of

functions termed quasi-concave functions

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Concave and

Quasi-Concave Functions

• The differences between concave and quasi-concave functions can be

illustrated with the function y = f(x1,x2) = (x1x2)k

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Concave and

Quasi-Concave Functions

• No matter what value k takes, this function is quasi-concave

• Whether or not the function is concave depends on the value of k

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Homogeneous Functions

• A function f(x₁,x₂,…x_n) is said to be homogeneous of degree k if

f(tx₁,tx₂,…tx_n) = tk f(x₁,x₂,…x_n)

– when a function is homogeneous of degree one, a doubling of all of its arguments

doubles the value of the function itself

– when a function is homogeneous of degree zero, a doubling of all of its arguments

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Homogeneous Functions

• If a function is homogeneous of degree

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Euler’s Theorem

• If we differentiate the definition for homogeneity with respect to the proportionality factor t, we get

ktk-1f(x₁,…,x_n) = x₁f₁(tx₁,…,tx_n) + … + x_nf_n(x₁,…,x_n)

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Euler’s Theorem

• Euler’s theorem shows that, for

homogeneous functions, there is a definite relationship between the

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Homothetic Functions

• A homothetic function is one that is formed by taking a monotonic

transformation of a homogeneous function

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Homothetic Functions

• For both homogeneous and homothetic functions, the implicit trade-offs among the variables in the function depend

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• Suppose we are examining the simple, two variable implicit function f(x,y) = 0

• The implicit trade-off between x and y for a two-variable function is

dy/dx = -fx/fy

• If we assume f is homogeneous of

degree k, its partial derivatives will be

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Homothetic Functions

• The implicit trade-off between x and y is

) , ( ) , ( ) , ( ) , ( 1 1 ty tx f ty tx f ty tx f t ty tx f t dx dy y x y k x k     _ 

• If t = 1/y,

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• The trade-off is unaffected by the

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Important Points to Note:

• Using mathematics provides a convenient, short-hand way for

economists to develop their models

– implications of various economic assumptions can be studied in a

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Important Points to Note:

• Derivatives are often used in economics because economists are interested in

how marginal changes in one variable affect another

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• The mathematics of optimization is an important tool for the development of models that assume that economic

agents rationally pursue some goal

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Important Points to Note:

• Most economic optimization

problems involve constraints on the choices that agents can make

– the first-order conditions for a

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• The Lagrangian multiplier is used to help solve constrained maximization problems

– the Lagrangian multiplier can be

(107)

Important Points to Note:

• The implicit function theorem illustrates the dependence of the choices that

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• The envelope theorem examines how optimal choices will change as the problem’s parameters change • Some optimization problems may

involve constraints that are

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• First-order conditions are necessary but not sufficient for ensuring a

maximum or minimum

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Important Points to Note:

• Certain types of functions occur in many economic problems

– quasi-concave functions obey the

second-order conditions of constrained maximum or minimum problems when the constraints are linear

– homothetic functions have the property that implicit trade-offs among the