```RS - Ch 9 - Optimization: One Variable
Chapter 9
Optimization: One Choice Variable
1
Léon Walras (1834-1910)
9.1
Vilfredo Federico D. Pareto (1848–1923)
Optimum Values and Extreme Values
• Goal vs. non-goal equilibrium
• In the optimization process, we need to identify the objective
function to optimize.
• In the objective function the dependent variable represents the
object of maximization or minimization
Example:
- Define profit function:
- Objective: Maximize π
- Tool: Q
π = PQ - C(Q)
2
RS - Ch 9 - Optimization: One Variable
9.2 Relative Maximum and Minimum:
First-Derivative Test
• Critical Value
The critical value of x is the value x0 if f’(x0) = 0
• A stationary value of y is f(x0)
• A stationary point is the point with coordinates x0 and f(x0)
• A stationary point is coordinate of the extremum
• Theorem (Weierstrass)
Let f : S→R be a real-valued function defined on a compact
(bounded and closed) set S ∈ Rn. If f is continuous on S, then f
attains its maximum and minimum values on S. That is, there
exists a point c1 and c2 such that
3
f (c1) ≤ f (x) ≤ f (c2)
for all x ∈ S.
9.2 First-derivative test
☺
• The first-order condition (f.o.c.) or necessary
condition for extrema is that f '(x*) = 0 and
the value of f(x*) is:
• A relative minimum if f '(x*) changes its sign
from negative to positive from the
immediate left of x0 to its immediate right.
(first derivative test of min.) ☺
y
B
f '(x*)=0
x
x*
• A relative maximum if the derivative f '(x)
changes its sign from positive to negative
from the immediate left of the point x* to
its immediate right. (first derivative test for a
max.) y
A
f '(x*) = 0
x*
4
RS - Ch 9 - Optimization: One Variable
9.2 First-derivative test
• The first-order condition or necessary condition for extrema is
that f '(x*) = 0 and the value of f(x*) is:
• Neither a relative maxima nor a relative minima if
f '(x) has the same sign on both the immediate left and right of
point x0. (first derivative test for point of inflection)
y
D
•
f '(x*) = 0
x
x*
5
9.2 Example: Average Cost Function
AC = Q 2 − 5Q + 8
Objective function
f (Q ) = 2Q − 5
1st derivative function
f / (Q ) = 2Q − 5 = 0
f.o.c.
/
*
Q = 5 / 2 = 2.5
left
extrema
( )
AC = f Q*
right
f (2.4 ) = 1.76
f (2.5) = 1.75
f (2.6 ) = 1.76 relative min
f ' (2.4) = −0.2
f / (2.5) = 0
f ' (2.6) = 0.2
(−,+ )
6
RS - Ch 9 - Optimization: One Variable
9.2 Example: Average Cost, 1st & 2nd Derivatives
1)
AC = Q 2 − 5Q + 8
2)
MC = 2Q − 5
MC = 2Q − 5 = 0
Q=5 2
3)
MC ′ = 2
7
9.3 Second and Higher Derivatives (Revisited)
Derivative of a derivative for y=f (x)
• The first derivative f '(x) or dy/dx is itself a function of x, it
should be differentiable with respect to x, provided that it is
continuous and smooth.
• The result of this differentiation is known as the second
derivative of the function f and is denoted as f ''(x) or d2y/dx2.
• The second derivative can be differentiated with respect to x
again to produce a third derivative,
f '''(x) and so on to f(n)(x) or dny/dxn
• Example: Revenue and Marginal Revenue functions
1)
2)
3)
4)
R = f (Q ) = 1200 Q − 2 Q 2
f ′( Q ) = 1200 − 4 Q
f ′′(Q ) = − 4
f ′′′(Q ) = 0
primitive
function
1st derivative
2 nd derivative
3 rd derivative
8
RS - Ch 9 - Optimization: One Variable
9.3 Example: Revenue, 1st & 2nd Derivatives
1)
R = 1200Q − 2Q 2
2)
MR = 1200 − 4Q
1200 − 4Q = 0
Q = 300
3)
MR′ = −4
9
9.3 Interpretation of the second derivative
• f '(x) measures the rate of change of a function
– e.g., whether the slope is increasing or decreasing
• f ''(x) measures the rate of change in the rate of change of a
function
– e.g., whether the slope is increasing or decreasing at an
increasing or decreasing rate
– how the curve tends to bend itself
10
RS - Ch 9 - Optimization: One Variable
9.3 The smile test
☺
• Convexity and Concavity: The smile test for aximum/minimum
• If f "(x) < 0 for all x, then strictly concave.
=> critical points are global maxima
• If f "(x) > 0 for all x, then strictly convex.
=> critical points are global minima
☺
• If a concave utility function (typical for risk aversion) is assumed
for a utility maximizing representative agent, there is no need to
check for s.o.c. Similar situation for a concave production function
11
9.3 Example: Revenue Function
1)
TR = 1200Q − 2Q 2
Revenue function
2)
MR = 1200 − 4Q
1st derivative
2' )
MR = 1200 − 4Q* = 0 f.o.c.
3)
4)
Q* = 300
MR′ = −4
MR ′ < 0
5)
extrema
2nd derivative
maximum
12
RS - Ch 9 - Optimization: One Variable
9.3 Inflection Point
• Definition
A twice differentiable function f(x) has an inflection point at x iff the
second derivative of f(.) changes from negative (positive) in some
interval (m,x~) to positive (negative) in some interval (x~,m), where
x~∈(m,n).
• Alternative Definition
An inflection point is a point (x, y) on a function, f(x), at which the
first derivative, f′(x), is at an extremun, -i.e. a minimum or
maximum. (Note: f’’(x)=0 is necessary, but not sufficient condition.)
Example: U(w) = w -2 w2 + w3
U’(w)=4w +3 w2
U’’(w)=4+6w
=> 4+6w~ =0
13
w~=2/3 is an inflection point
9.3 Examples: Optimal Seignorage
e
M
= e − λ (π + r ) +α Y
P
e
M
S =π
= π e − λ (π + r ) +α Y
P
F.o.c. (assume π = π e ) :
Demand for money
Seignorage
dS
= e − λ (π + r ) +α Y + π ( − λ ) e − λ (π + r ) +α Y
dπ
= e − λ ( π + r ) + α Y + π ( − λ ) e − λ ( π + r ) + α Y = e − λ ( π + r ) + α Y (1 − πλ )
1
(1 − π * λ ) = 0 => π * =
(Critical point)
λ
S.o.c. :
d 2S
= − λ e − λ ( π + r ) + α Y (1 − πλ ) + ( − λ ) e − λ ( π + r ) + α Y = − λ e − λ ( π + r ) + α Y ( 2 − πλ )
dπ 2
*
d 2S
1
1
( π * = ) = − λ e − λ ( π + r ) + α Y < 0 => π * =
is a maximum
2
dπ
λ
λ
RS - Ch 9 - Optimization: One Variable
9.3 Examples: Optimal Timing - wine storage
A(t ) = Ve − rt
V = ke
Present va lue
t
Growth in value
12
A(t) = ke t e − rt = ke t
ln A(t )
= lnk + lne t
(
= ln k + (t
12
½
− rt
− rt
)
− rt )
= lnk + t ½ − rt lne
dA 1
dt A
F.o.c. :
½
1 −1 2
t
−r
2
dA
1 −1 2
 1 −1 2

= A t * − r  = 0
=> t * − r = 0
dt
2
2

1
1
=>
=r
=> t* = 2
4r
2 t*
=
9.3 Examples: Optimal Timing - wine storage
Optimal time : t * =
1
4r 2
Let ( r ) = 10 %
1
= 25 years
4( 0 .10 )2
Determine optimal values for A(t) and V :
t* =
A(t) = ke t
½
− rt
let ( k ) = \$ 1 / bottle
A(t) = e 5 −(.1 )(( 25 ) = e 2 .5 = \$ 12 .18 / bottle
V = Ae rt
V = ( \$12 .18 /bottle) e (.1 )( 25 ) = \$ 148 .38 / bottle
V = \$ 148 .38 / bottle
16
RS - Ch 9 - Optimization: One Variable
9.3 Examples: Optimal Timing - wine storage
Plot of .5(t).5(t)-.5=r, r=.10, t=25
17
9.4 Formal Second-Derivative Test: Necessary
and Sufficient Conditions
• The zero slope condition is a necessary condition and since it is
found with the first derivative, we refer to it as a 1st order
condition.
• The sign of the second derivative is sufficient to establish the
stationary value in question as a relative minimum if f "(x0) >0,
the 2nd order condition or relative maximum if f "(x0)<0.
18
RS - Ch 9 - Optimization: One Variable
9.4 Example 1: Profit function
Revenue and Cost functions
1)
TR = 1200Q − 2Q 2
2)
TC = Q 3 − 61.25Q 2 + 1528 .5Q + 2000
Profit function
3)
π = TR-TC = − Q 3 + 59.25Q 2 − 328.5Q − 2000
1st derivative of profit function
4)
π′ = −3Q 2 + 118.5Q − 328 .5 = 0
5)
Q1* = 3
Q2* = 36.5
2 nd derivative of profit function
6) π′′ = −6Q + 118.5
7 ) π′′(3) = 100.5
π′′(36.5) = −100.5
applying the smile test
8)
( )
( )
π′′ Q1* > 0 → min π′′ Q2* < 0 → max
19
9.4 Example 2: Imperfect Competition
Total revenue function
1)
(
)
TR ≡ AR * Q = 8000 − 23 Q + 1 . 1Q 2 − 0 . 018 Q 3 Q
20
RS - Ch 9 - Optimization: One Variable
9.4 Example 2: TR and 1st , 2nd & 3rd derivatives
Average
1)
and total revenue functions
AR = f (Q ) = 8000 − 23 Q + 1 . 1Q 2 − 0 . 018 Q 3
2)
TR = f (Q )Q
Marginal revenue and the product
MR = f ( Q ) Q ′ + Q f ′ ( Q )
3)
4)
rule
(
)
M R = 8000 − 23 Q + 1 . 1Q 2 − 0 . 018 Q 3 (1 )
(
2
+ Q − 23 + 2 . 2 Q − . 054 Q
)
= 8000 − 46 Q + 3 . 3 Q 2 − 0 . 0 . 72 Q 3
2 nd derivative
5)
M R ′ = f ′ (MR )
6)
M R ′ = − 46 + 6 . 6 Q − 0 . 216 Q 2 = 0
7)
Q 1* = 10 . 76
Q 2* = 19 . 79
3rd derivitive
M R ′′ = f ′ (M R ′ )
M R ′′ ( Q ) = 6 . 6 − 0 . 432 Q
M R ′′ (10 . 76 ) = 1 . 95
MR " (19 . 79 ) = − 1 . 94
8)
9)
10 )
smile test
11 )
M R ′′ ( Q 1* ) > 0 → min.
21
MR " ( Q 2* ) < 0 → max.
9.5 Taylor Series of a polynomial function:
Revisited
Taylor series for an arbitrary function : Any function can be approximat ed
by the weighted sum of its derivative s. Then the change is given by
Where R n +1
/
f
f ( x ) − f (x0 ) =
(x0 ) (x − x )1
0
1!
f
+
//
(x0 ) (x − x )2
0
2!
+…+
f (n) (x 0 )
(x − x0 )n + R n +1
n!
f (n +1 ) ( p )
(x − x0 )n +1
=
(n + 1)!
If n = 2, then
f(x) - f(x 0 ) =
f
/
(x0 ) (x − x )1
0
1!
At x 0 , f / (x 0 ) = 0
(x0 ) (x − x )2 + R
0
3
2!
- i.e., x 0 is a max., min., or inflection .
+
f
//
Then,
f(x) - f(x 0 ) ≈
f
(x0 ) (x − x )2
0
2!
//
=> the sign of f // (x 0 ) determines what x 0 is.
22
RS - Ch 9 - Optimization: One Variable
9.5 Taylor expansion and relative extremum
A function f(x) attains a relative max (min) value at x 0 if f(x) -f(x 0 ) is neg. (pos.)
for values of x in the immediate neighborho od of x 0 (the critical value) both to its
left and right Taylor series approximat ion for a small change in x :
(x0 ) (x − x )2 + ... + R
0
n +1
1!
2!
At the max., min., or inflection , f / (x 0 ) = 0
f(x) - f(x 0 ) =
f
/
(x0 ) (x − x )1
0
and if f // (x 0 ) = 0, and if f
(3)
+
f
//
(x 0 ) = 0, L
then f(x) - f(x 0 ) = R n +1
What is the sign of R for the first nonzero derivative ?
23
9.5 Nth-derivative test
If the first derivative of a function f(x) at x0 is f
/
(x0 ) =
0 and
if the first nonzero derivative value at x 0 encountered in
successive derivation is that of the N th derivative, f (n)(x0 ) ≠ 0,
then the stationary value f ( x0 ) will be :
a relative max if N is even and f (n)(x0 ) < 0
a relative min if N is even and f (n)(x0 ) > 0
an inflection point if N is odd
24
RS - Ch 9 - Optimization: One Variable
9.5 Nth-derivative test
Y = ( 7-x)4
primitive function
Y / = -4( 7-x)3
1st deriative
Y / = 0 at x * = 7
the critical value
Y // (7) = 12( 7-x)2 = 0
2 nd deriviative
Y ( 3) (7) = -24( 7-x) = 0
3rd derivative
Y ( 4 ) (7) = 24
4 th derivative
Because first nonzero derivative Y (n ) is even (4)
and Y (4) > 0 (24), critical value is a min.
25
9.5 Nth-derivative test
Y = ( 7-x)4
primitive function
M
Y ( 4) = 24
4 th derivative
decision rule :
n is even (4) and > 0
therefore a minimum
26
```