ECON 1150, Spring 2013 Lecture 3: Optimization: One Choice Variable Necessary conditions Sufficient...

ECON 1150, Spring 2013

Lecture 3: Optimization:One Choice Variable

Necessary conditionsSufficient conditions

Reference:

Jacques, Chapter 4

Sydsaeter and Hammond, Chapter 8.


1. Optimization Problems

Economic problems

Consumers: Utility maximization

Producers: Profit maximization

Government: Welfare maximization


Maximization problem: maxx f(x)

f(x): Objective function with a domain D

x: Choice variable

x*: Solution of the maximization problem

A function defined on D has a maximum point at x* if

f(x) f(x*) for all x D.

f(x*) is called the maximum value of the function.


Minimization problem: minx f(x)

f(x): Objective function with a domain D

x: Choice variable

x*: Solution of the maximization problem

A function defined on D has a minimum point at x* if

f(x) f(x*) for all x D.

f(x*) is called the minimum value of the function.


Example 3.1: Find possible maximum and minimum points for:

a. f(x) = 3 – (x – 2)2;

b. g(x) = (x – 5) – 100, x 5.


2. Necessary Condition for Extrema

What are the maximum and minimum points of the following functions?

y = 60x – 0.2x2

y = x3 – 12x2 + 36x + 8


x

y

0 x*

y*

x1 x2

Maximum

x < x* : dy / dx > 0

x > x* : dy / dx < 0

x = x* : dy / dx = 0

Characteristic of a maximum point


x

y

0 x*

y*

x1 x2

Minimum

x < x* : dy / dx < 0

x > x* : dy / dx > 0

x = x* : dy / dx = 0

Characteristic of a minimum point


Theorem: (First-order condition for an extremum) Let y = f(x) be a differentiable function. If the function achieves a maximum or a minimum at the point x = x*, then

dy / dx |x=x* = f’(x*) = 0

Stationary point : x*

Stationary value : y* = f(x*)


Example 3.2: Find the stationary point of the function y = 60x – 0.2x2.

The first-order condition is a necessary, but not sufficient, condition.

x

y

-50 0 50 100 150 200 250 300

-1000

1000

2000

3000

4000


Example 3.3: Find the stationary values of the function y = f(x) = x3 – 12x2 + 36x + 8.

x

y

-1 0 1 2 3 4 5 6 7 8 9 10

-60

-40

-20

20

40

60

80

100


3. Finding Global Extreme Points

Possibilities of the nature of a function f(x) at x = c.

• f is differentiable at c and c is an interior point.

• f is differentiable at c and c is a boundary point.

• f is not differentiable at c.


3.1 Simple Method

a. Find all stationary points of f(x) in (a,b)

b. Evaluate f(x) at the end points a and d and at all stationary points

c. The largest function value in (b) is the global maximum value in [a,b].

d. The smallest function value in (b) is the global minimum value in [a,b].

Consider a differentiable function f(x) in [a,b].


3.2 First-Derivative Test for Global Extreme Points

x

y

0 x*

y*

x1 x2 x

y

0 x*

y*

x1 x2

Global maximum Global minimum


• If f’(x) 0 for x c and f’(x) 0 for x c, then x = c is a global maximum point for f.

• If f’(x) 0 for x c and f’(x) 0 for x c, then x = c is a global minimum point for f.

First-derivative Test


Example 3.4: Consider the function

y = 60x – 0.2x2.

a. Find f’(x).

b. Find the intervals where f increases and decreases and determine possible extreme points and values.


Example 3.5: y = f(x) = e2x – 5ex + 4.

a. Find f’(x).


c. Examine limx f(x) and limx-f(x).


3.3 Extreme Points for Concave and Convex Functions

Let c be a stationary point for f.

a. If f is a concave function, then c is a global maximum point for f.

b. If f is a convex function, then c is a global minimum point for f.


Example 3.6: Show that f(x) = ex–1 – x. is a convex function and find its global minimum point.

Example 3.7: The profit function of a firm is (Q) = -19.068 + 1.1976Q – 0.07Q1.5. Find the value of Q that maximizes profits.


4. Identifying Local Extreme Points

0

x

y

a

b

c

d


Let a < c < b.

a. If f’(x) 0 for a < x < c and f’(x) 0 for c < x < b, then x = c is a local maximum point for f.

b. If f’(x) 0 for a < x < c and f’(x) 0 for c < x < b, then x = c is a local minimum point for f.

4.1 First-derivative Test for Local Extreme Points


Example 3.8: y = f(x) = x3 – 12x2 + 36x + 8.

a. Find f’(x).


c. Examine limx f(x) and limx-f(x).


Example 3.9: Classify the stationary points of the following functions.

.exf(x) .b

;1x3

2x

6

1x

9

1)x(f .a

x2

23


4.2 Second-Derivative Test

The nature of a stationary point:

Decreasing slope Local maximum

x

y

0 x*

y*

x1 x2

x

dy/dx

y*

0 x*x1x2

0

dx

sloped



Increasing slope Local minimum

dy/dx

x0 x*x1 x2

x

y

0 x*

y*

x1 x20

dx

)slope(d



Point of inflection Stationary slope

0dx

)slope(d


Second order condition: Let y = f(x) be a differentiable function and f’(c) = 0.

f”(c) < 0 Local maximum

f”(c) > 0 Local minimum

f”(c) = 0 No conclusion


Example 3.10: Identify the nature of the stationary points of the following functions:

a. y = 4x2 – 5x + 10;

b. y = x3 – 3x2 + 2;

c. y = 0.5x4 – 3x3 + 2x2.


0X

Y

0X

Y

4.3 Point of Inflection

a b


Let f be a twice differentiable function.

a. If c is an inflection point for f, then f”(c) = 0.

b. If f”(c) = 0 and f” changes sign around c, then c is an inflection point for f.

Test for inflection points:


x

y

-2 -1 0 1 2 3

-10

10

20

30

Example 3.11: y = 16x – 4x3 + x4. dy / dx = 16 – 12x2 + 4x3.At x = 2, dy/dx = 0. However, the point at x = 2 is neither a maximum nor a minimum.

Point of inflection


Example 3.12: Find possible inflection points for the following functions,

a. f(x) = x6 – 10x4.

b. f(x) = x4.


4.4 From Local to Global

a. Find all local maximum points of f(x) in [a,b]

b. Evaluate f(x) at the end points a and b and at all local maximum points

c. The largest function value in (b) is the global maximum value in [a,b].

Consider a differentiable function f(x) in [a,b].


5. Curve Sketching

• Find the domain of the function

• Find the x- and y- intercepts

• Locate stationary points and values

• Classify stationary points

• Locate other points of inflection, if any

• Show behavior near points where the function is not defined

• Show behavior as x tends to positive and negative infinity


Example 3.13: Sketch the graphs of the following functions by hand, analyzing all important features.

a. y = x3 – 12x;

b. y = (x – 3)x;

c. y = (1/x) – (1/x2).


6. Profit Maximization

Total revenue: TR(q) Marginal revenue: MR(q)

Total cost: TC(q) Marginal cost: MC(q)

Profit: (q) = TR(q) – TC(q)

Principles of Economics:

MC = MR

MC curve cuts MR curve from below.


Calculus

First-order condition:

I.e., MR – MC = 0

Thus the marginal condition for profit maximization is just the first-order condition.

0dq

dTC

dq

dTR

dq

d


Second-order condition:

Calculus

At profit-maximization, the slope of the MR curve is smaller than the slope of the MC curve.

0dq

dMC

dq

dMR

dq

TCd

dq

TRd

dq

d2

2

2

2

2

2

dq

dMC

dq

dMR


6.1 A Competitive Firm

Example 3.14: Given (a) perfect competition; (b) market price p; (c) the total cost of a firm is TC(q) = 0.5q3– 2q2 + 3q + 2. If p = 3, find the maximum profit of the firm.

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ECON 1150, Spring 2013 Lecture 3: Optimization: One Choice Variable Necessary conditions Sufficient...

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