Lecture 3: Optimization: One Choice Variable

Necessary conditions Sufficient conditions

Reference: Jacques, Chapter 4 Sydsaeter and Hammond, Chapter 8.

ECON 1150, Spring 2013

1. Optimization Problems

Economic problems Consumers: Utility maximization Producers: Profit maximization Government: Welfare maximization ECON 1150, Spring 2013

Maximization problem: max x 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 ECON 1150, Spring 2013 function.

Minimization problem: min x 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 ECON 1150, Spring 2013 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.

ECON 1150, Spring 2013

2. Necessary Condition for Extrema

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

y = 60x – 0.2x

2 y = x 3 – 12x 2 + 36x + 8 ECON 1150, Spring 2013

Characteristic of a maximum point y* y Maximum x < x* : dy / dx > 0 x > x* : dy / dx < 0 x = x* : dy / dx = 0 0 x 1 x* x 2 x ECON 1150, Spring 2013

y* Characteristic of a minimum point y Minimum x < x* : dy / dx < 0 x > x* : dy / dx > 0 x = x* : dy / dx = 0 0 x 1 x* x x 2 ECON 1150, Spring 2013

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*) ECON 1150, Spring 2013

Example 3.2

: Find the stationary point of the function y = 60x – 0.2x

2 .

y

4000 3000 2000 1000 -50 -1000 0 50 100 150 200 250 300

x

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

Example 3.3

: Find the stationary values of the function y = f(x) = x 3 – 12x 2 + 36x + 8.

100 80 60 40 20

y

-1 -20 0 -40 -60 1 2 3 4 5 6 7 8 9 10

x

ECON 1150, Spring 2013

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.

ECON 1150, Spring 2013

3.1 Simple Method

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

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].

ECON 1150, Spring 2013

3.2 First-Derivative Test for Global Extreme Points

Global maximum y y Global minimum y* y* 0 x 1 x* x 2 x 0 x 1 ECON 1150, Spring 2013 x* x 2 x

First-derivative Test • 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.

ECON 1150, Spring 2013

Example 3.4

: Consider the function y = 60x – 0.2x

2 .

a. Find f’(x).

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

ECON 1150, Spring 2013

Example 3.5

: y = f(x) = e 2x – 5e x + 4.

a. Find f’(x).

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

c. Examine lim x  f(x) and lim x   f(x).

ECON 1150, Spring 2013

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

Example 3.6

: Show that f(x) = e x –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 – 0.07Q

1.5

. Find the value of Q that maximizes profits.

ECON 1150, Spring 2013

0

4. Identifying Local Extreme Points

y c a d b x ECON 1150, Spring 2013

4.1 First-derivative Test for Local Extreme Points

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.

ECON 1150, Spring 2013

Example 3.8

: y = f(x) = x 3 + 8.

– 12x 2 + 36x a. Find f’(x).

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

c. Examine lim x  f(x) and lim x   f(x).

ECON 1150, Spring 2013

Example 3.9

: Classify the stationary points of the following functions.

a b .

.

f ( x f(x) )  1 9 x 3  x 2 e x .

 1 6 x 2  2 3 x  1 ; ECON 1150, Spring 2013

4.2 Second-Derivative Test

The nature of a stationary point: Decreasing slope  Local maximum y dy/dx d  slope   0 dx y* y* 0 x 1 x* x 2 x 0 x 1 x* x 2 x ECON 1150, Spring 2013

The nature of a stationary point: Increasing slope  Local minimum y dy/dx y* 0 x 1 x* 0 x 1 x 2 x ECON 1150, Spring 2013 x* x x 2 d ( slope )  0 dx

The nature of a stationary point: Point of inflection  Stationary slope

d ( slope )



0 dx

ECON 1150, Spring 2013

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 ECON 1150, Spring 2013

Example 3.10

: Identify the nature of the stationary points of the following functions: a. y = 4x 2 – 5x + 10; b. y = x 3 – 3x 2 + 2; c. y = 0.5x

4 – 3x 3 + 2x 2 .

ECON 1150, Spring 2013

Y

4.3 Point of Inflection

Y 0 a X 0 ECON 1150, Spring 2013 b X

Test for inflection points: 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.

ECON 1150, Spring 2013

Example 3.11

: y = 16x – 4x 3 dy / dx = 16 – 12x 2 + 4x 3 .

+ x 4 .

At x = 2, dy/dx = 0. However, the point at x = 2 is neither a maximum nor a minimum.

y

30 20 -2 10 -1 -10 0 1 2 3 ECON 1150, Spring 2013

x

Point of inflection

Example 3.12

: Find possible inflection points for the following functions, a. f(x) = x 6 – 10x 4 .

b. f(x) = x 4 .

ECON 1150, Spring 2013

4.4 From Local to Global

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

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].

ECON 1150, Spring 2013

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 ECON 1150, Spring 2013

Example 3.13

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

a. y = x 3 – 12x; b. y = (x – 3)  x; c. y = (1/x) – (1/x 2 ).

ECON 1150, Spring 2013

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 .

ECON 1150, Spring 2013

Calculus First-order condition: d  dTR  dTC dq  dq dq  0 I.e., MR – MC = 0 Thus the marginal condition for profit maximization is just the first-order condition.

ECON 1150, Spring 2013

Calculus Second-order condition: d 2  dq 2  d 2 TR dq 2  d 2 TC dq 2  dMR dq  dMC dq  0 dMR dq  dMC dq At profit-maximization, the slope of the MR curve is smaller than the slope of the MC curve. ECON 1150, Spring 2013

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) = 0.5q

3 – 2q 2 + 3q + 2. If p = 3, find the maximum profit of the firm.

ECON 1150, Spring 2013