1 / 39

Lecture 3: Optimization : One Choice Variable

Lecture 3: Optimization : One Choice Variable. Necessary condition s Sufficient condition s Reference: Jacques, Chapter 4 Sydsaeter and Hammond, Chapter 8. 1. Optimization Problems. Economic problems Consumers: Utility maximization Producers: Profit maximization

sheryl
Télécharger la présentation

Lecture 3: Optimization : One Choice Variable

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 3: Optimization:One Choice Variable Necessary conditions Sufficient conditions Reference: Jacques, Chapter 4 Sydsaeter and Hammond, Chapter 8. ECON 1150, Spring 2013

  2. 1. Optimization Problems Economic problems Consumers: Utility maximization Producers: Profit maximization Government: Welfare maximization ECON 1150, Spring 2013

  3. 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. ECON 1150, Spring 2013

  4. 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. ECON 1150, Spring 2013

  5. Example 3.1: Find possible maximum and minimum points for: • f(x) = 3 – (x – 2)2; • g(x) = (x – 5) – 100, x  5. ECON 1150, Spring 2013

  6. 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 ECON 1150, Spring 2013

  7. y y* x 0 x1 x* x2 Characteristic of a maximum point Maximum x < x* : dy / dx > 0 x > x* : dy / dx < 0 x = x* : dy / dx = 0 ECON 1150, Spring 2013

  8. y y* x 0 x1 x* x2 Characteristic of a minimum point Minimum x < x* : dy / dx < 0 x > x* : dy / dx > 0 x = x* : dy / dx = 0 ECON 1150, Spring 2013

  9. 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

  10. 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. ECON 1150, Spring 2013

  11. Example 3.3: Find the stationary values of the function y = f(x) = x3 – 12x2 + 36x + 8. ECON 1150, Spring 2013

  12. 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

  13. 3.1 Simple Method • Find all stationary points of f(x) in (a,b) • Evaluate f(x) at the end points a and d and at all stationary points • The largest function value in (b) is the global maximum value in [a,b]. • The smallest function value in (b) is the global minimum value in [a,b]. Consider a differentiable function f(x) in [a,b]. ECON 1150, Spring 2013

  14. y y y* y* x x 0 0 x1 x1 x* x* x2 x2 3.2 First-Derivative Test for Global Extreme Points Global maximum Global minimum ECON 1150, Spring 2013

  15. 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

  16. 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. ECON 1150, Spring 2013

  17. Example 3.5: y = f(x) = e2x – 5ex + 4. a. Find f’(x). b. Find the intervals where f increases and decreases and determine possible extreme points and values. c. Examine limx f(x) and limx-f(x). ECON 1150, Spring 2013

  18. 3.3 Extreme Points for Concave and Convex Functions Let c be a stationary point for f. • If f is a concave function, then c is a global maximum point for f. • If f is a convex function, then c is a global minimum point for f. ECON 1150, Spring 2013

  19. 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. ECON 1150, Spring 2013

  20. y c a d b x 0 4. Identifying Local Extreme Points ECON 1150, Spring 2013

  21. 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

  22. Example 3.8: y = f(x) = x3 – 12x2 + 36x + 8. a. Find f’(x). b. Find the intervals where f increases and decreases and determine possible extreme points and values. c. Examine limx f(x) and limx-f(x). ECON 1150, Spring 2013

  23. Example 3.9: Classify the stationary points of the following functions. ECON 1150, Spring 2013

  24. y dy/dx y* y* x2 x x* 0 x1 x 0 x1 x* x2 4.2 Second-Derivative Test The nature of a stationary point: Decreasing slope  Local maximum ECON 1150, Spring 2013

  25. y dy/dx y* x x1 x* x2 0 x 0 x1 x* x2 The nature of a stationary point: Increasing slope  Local minimum ECON 1150, Spring 2013

  26. The nature of a stationary point: Point of inflection  Stationary slope ECON 1150, Spring 2013

  27. 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

  28. Example 3.10: Identify the nature of the stationary points of the following functions: • y = 4x2 – 5x + 10; • y = x3 – 3x2 + 2; • y = 0.5x4 – 3x3 + 2x2. ECON 1150, Spring 2013

  29. Y Y X X 0 0 4.3 Point of Inflection a b ECON 1150, Spring 2013

  30. 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

  31. Point of inflection 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. ECON 1150, Spring 2013

  32. Example 3.12: Find possible inflection points for the following functions, a. f(x) = x6 – 10x4. b. f(x) = x4. ECON 1150, Spring 2013

  33. 4.4 From Local to Global • Find all local maximum points of f(x) in [a,b] • Evaluate f(x) at the end points a and b and at all local maximum points • The largest function value in (b) is the global maximum value in [a,b]. Consider a differentiable function f(x) in [a,b]. ECON 1150, Spring 2013

  34. 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

  35. Example 3.13: Sketch the graphs of the following functions by hand, analyzing all important features. • a. y = x3 – 12x; • y = (x – 3)x; • y = (1/x) – (1/x2). ECON 1150, Spring 2013

  36. 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

  37. Calculus First-order condition: I.e., MR – MC = 0 Thus the marginal condition for profit maximization is just the first-order condition. ECON 1150, Spring 2013

  38. Calculus Second-order condition: At profit-maximization, the slope of the MR curve is smaller than the slope of the MC curve. ECON 1150, Spring 2013

  39. 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. ECON 1150, Spring 2013

More Related