lecture 3: optimization : one choice variable
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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 - PowerPoint PPT PresentationTRANSCRIPT
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ECON 1150, Spring 2013
Lecture 3: Optimization:One Choice Variable
Necessary conditionsSufficient conditions
Reference:
Jacques, Chapter 4
Sydsaeter and Hammond, Chapter 8.
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ECON 1150, Spring 2013
1. Optimization Problems
Economic problems
Consumers: Utility maximization
Producers: Profit maximization
Government: Welfare maximization
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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
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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*)
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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
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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.
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ECON 1150, Spring 2013
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].
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
• 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
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
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).
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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 global minimum point for f.
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
4. Identifying Local Extreme Points
0
x
y
a
b
c
d
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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).
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
The nature of a stationary point:
Increasing slope Local minimum
dy/dx
x0 x*x1 x2
x
y
0 x*
y*
x1 x20
dx
)slope(d
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ECON 1150, Spring 2013
The nature of a stationary point:
Point of inflection Stationary slope
0dx
)slope(d
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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
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ECON 1150, Spring 2013
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.
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ECON 1150, Spring 2013
0X
Y
0X
Y
4.3 Point of Inflection
a b
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ECON 1150, Spring 2013
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:
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
Example 3.12: Find possible inflection points for the following functions,
a. f(x) = x6 – 10x4.
b. f(x) = x4.
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ECON 1150, Spring 2013
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].
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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
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ECON 1150, Spring 2013
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).
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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.
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ECON 1150, Spring 2013
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
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ECON 1150, Spring 2013
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
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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) = 0.5q3– 2q2 + 3q + 2. If p = 3, find the maximum profit of the firm.