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THE MATHEMATICS OF OPTIMIZATION

Copyright ©2005 by South-Western, a division of Thomson Learning. All rights reserved. Walter Nicholson, Microeconomic Theory - Basic Principles and Extensions, Thompson-Southwestern, 9th Edition, 2005, ISBN: 0324270860

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Maximization of a Function of One Variable

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

)(qf

= f(q)

Quantity

*

q*

Maximum profits of* occur at 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)

Quantity

*

q*

1

q1

2

q2

0q

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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)

Quantity

*

q*

0q

3

q3

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Basic Differentiation Rules

1.

Ex.

2.

0 is a constantdc c

dx

( ) 5

( ) 0

f x

f x

Ex.

1 is a real numbern ndx nx n

dx

7

6

( )

( ) 7

f x x

f x x

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3. Sum-Difference Rule. If f(x) = g(x) ± h(x), then f’(x) = g’(x) ±

h’(x)

Define then

Thus

( ) ) f x u v

'( ) ( )d du dv

f x u vdx dx dx

( ) u g x ( ) v h x

4. Product Rule: If f(x) = g(x)h(x), then f’(x) = g(x)h’(x) +

h(x)g’(x)

Define then

'( ) ( )d dv du

f x uv u vdx dx dx

( ) u g x ( ) v h x ( ) ) f x uv

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5. Quotient Rule

If f(x) = then

2

[ ]'( )

du dvv udx dxf xv

( )

( )

g x

h x 2

( ) '( ) ( ) '( )'( )

( )

h x g x g x h xf x

h x

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Differentiation of Logarithmic Functions

2

2

ln 2

4 2'( )

2

y x

xf x

xx

1'( ) ln 0

df x x x

dx x

1'( ) ln

d duf x u

dx u dx

Generalized Rule for Natural Logarithm Functions

Derivative of the Natural Logarithm

If u is a differentiable function, then

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The Chain Rule

( )( ) ( )

d du df u duf u f u

dx dx du dx

The derivative of a f (quantity) is the derivative of f evaluated at the quantity, times the derivative of the quantity.

If f is a differentiable function of u and u is a differentiable function of x, then the composite f (u) is a differentiable function of x, and

( , )

( , )

TR PQ K L

dTR dQ K L dQP

dL dQ dL

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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

• Since the second derivative is always -10, q = 100 is a global maximum

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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

),...,,( nxxxfy 21

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• The partial derivative of y with respect to x1 is denoted by

Partial Derivatives

1

111

ffx

f

x

yx or or or

• It is understood that in calculating the partial derivative, all of the other x’s are held constant

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Calculating Partial Derivatives

2122

2111

2221

2121

2

2

cxbxfx

f

bxaxfx

f

cxxbxaxxxfy

and

then ,),( If 1.

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211

1

2121

and

then,lnln),( If 2.

x

bf

x

f

x

af

x

f

xbxaxxfy

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Total Differential

• Suppose that y = f(x1,x2,…,xn)

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

n

n

dxx

fdx

x

fdx

x

fdy

...2

2

1

1

nndxfdxfdxfdy ...2211

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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...21 nfff

• A point where this condition holds is called a critical point

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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 maximization problems is the Lagrangian multiplier method

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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,…, xn)

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

g(x1, x2,…, xn) = 0

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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 because g(x1, x2,…, xn) = 0

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Lagrangian Multiplier Method

• First-Order Conditions

L/x1 = f1 + g1 = 0

L/x2 = f2 + g2 = 0

.

L/xn = fn + gn = 0

.

.

L/ = g(x1, x2,…, xn) = 0

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Lagrangian Multiplier Method

• 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– these x’s will make the value of L (and

therefore f) as large as possible

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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

– the denominators reflect the added burden on the constraint of using more xi

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Lagrangian Multiplier Method

• At the optimal choices for the x’s, the ratio of the marginal benefit of increasing xi to the marginal cost of increasing xi 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

provides a “shadow price” to the constraint

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Lagrangian Multiplier Method

• 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

=0 implies that the constraint is not binding