Transcript Unconstrained Optimization
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
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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].
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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