Necessary conditions
Sufficient conditions
Reference:
Jacques, Chapter 4
Sydsaeter and Hammond, Chapter 8.
ECON 1150, Spring 2013

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

What are the maximum and minimum points of the following functions?
y = 60x – 0.2x
2 y = x 3 – 12x 2 + 36x + 8
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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.
-1
-20
0
-40
-60
100
80
60
40
20 y
1 2 3 4 5 6 7 8 9 10 x
ECON 1150, Spring 2013

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

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

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

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
y c a d b x
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.
ECON 1150, Spring 2013

Example 3.8
: y = f(x) = x 3 – 12x 2 + 36x
+ 8.
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

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

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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 .
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Y
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 + x 4 .
dy / dx = 16 – 12x 2 + 4x 3 .
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 .
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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

• 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

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
 dq
 dTR dq
 dTC 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

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