# Document

```Chapter 2
Mathematics of Optimization
•
Many economic concepts can be
expressed as functions (eg., cost, utility,
profit),
– To be optimized (maximized or minimized),
– Subject to or not subject to constraints (eg.,
income or resource constraints).
•
Optimizing a function of one variable
(unconstrained):
Maximization (one variable
and no constraint)
• To maximize π=f(q), look
at changes in π as q
increases.
• As long as Δ π &gt; 0, π
increases.
• Can express Δ’s as a
ratio ΔΠ
; could be + or -.
Δq
• Limit of ratio as Δq 0 is
derivative of π=f(q).
• For q&lt;q*, derivative &gt;0.
• For q&gt;q*, derivative &lt;0.
• At q*, derivative = 0.
Π
Π =f(q) is concave.
∆Π=0
Π=f(q)
∆Π&gt;0
∆Π&lt;0
∆q&gt;0
q*
∆q&gt;0
q
Minimization (one variable and
no constraint)
• To minimize AC=f(q),
look at ∆AC as q
increases.
• As long as ∆AC&lt;0, AC
is decreasing. The
change ratio is
AC =f(q) is convex.
AC
dAC
&lt;0
dq
dAC
&gt;0
dq
∆AC
∆q
Limit of ratio as ∆q
0 is
derivative of AC=f(q).
For q&lt;q**, derivative &lt; 0.
For q&gt;q**, derivative &gt; 0.
At q= q** , derivative = 0.
AC = f(q)
q**
q
• First derivative is the slope of a line tangent to the
Slope = dπ/dq at the point of tangency.
function at a point.
Π
• Derivative is denoted as:
dΠ
dAC
df
or
or
or f ′ (q) = f ′
dq
dq
dq
0
q
• To maximize or minimize f(q), find q value where:
df
= 0
dq
• Derivative will be a constant if the original function
is linear; thus, the function has no Max or Min.
• But if the function is nonlinear, its derivative will be
a function of q itself (the derivative varies as q
varies).
• For derivative at a specific value of q, say q1,
denote dΠ
At q*, dΠ
dq
q = q1
.
dq
q =q*
= 0.
• To find the q where Π is maximized or AC is
minimized, find the derivative of f(q), and set it
equal to 0 and solve for q to get q*.
• The “First Order Condition (FOC)” for a
maximum or a minimum of an unconstrained
function of one variable is:
df
FOC is
= 0.
dq
Solve the FOC for q to
get q* (the optimal q).
The FOC is a necessary condition for a maximum
(or minimum) but not a sufficient condition.
Second Order Condition (SOC)
The FOC is dy dx = 0.
Convex
function
y
0
x*
dy
= 0
dx
Concave function
y
0
x
y
0
x*
dy
= 0
dx
x
FOC is necessary but not sufficient
dy
dx
0
x
Increases
through 0
0
Convex then
concave
dyx*
=0
dx
x
0
x
x*
x*
Concave then
convex
dy
dx
Decreases
through 0
dy
dx
Inflection
point
x
x*
Take the derivative (second derivative) of the derivative (first derivative).
d2y
dx 2
+
0
d2y
dx 2
d2y
x*&gt; 0
dx 2
Min
x*
x
0
-
d2y
x*&lt; 0
dx 2
x*
Max
x
d2y
dx 2
Inflection
0
d2y
x* = 0
dx 2
x
x*
Examples:
y
dy
=0
dx
MAX
2
d y
d2y
= 0 dx 2 = 0
2
dx
d2y
=0
dx 2
Local maximum
and local minimum
dy
=0
dx
MIN
dy
&gt;0
dx
dy
&lt;0
dx
d2y
&lt;0
2
dx
x
dy
&gt;0
dx
d2y
&gt;0
2
dx
dy
&gt;0
dx
2
d2y
d
y
&lt;0
2
&gt;0
2
dx
dx
An inflection point is where the second derivative equals zero
when evaluated at the q* identified by the FOC.
When the first derivative equals zero, you have a minimum,
maximum, or an inflection point. Not all inflections have f ′ = 0.
SOC
if f ′′ q* &lt; 0, have a maximum at q*
if f ′′ * &gt; 0, have a minimum at q*
q
if f ′′ * = 0, have an inflection at q*
q
Max
FOC
f′ = 0
SOC
f ′′ q* &lt; 0
For an
unconstrained
function of
one variable
Min
f′ = 0
f ′′ q* &gt; 0 Solve FOC for q* and
evaluate f ′′ at q*.
Rules for Finding Derivatives
1.The derivative of a constant is 0. If b is a constant in
y = b, then
dy db
=
= 0.
dx dx
Parameters Variable
y = a + bx, the derivative of a with respect to x is 0.
2. The Power Rule – The derivative of a variable raise to
b
dax
dax
a power is
= bax b −1 ,
= ax1−1 = a ,
dx
dx
= 1,
dx
dx b
= bx b −1.
dx
dx
• Examples:
For y = a+bx2
dy
= 2bx
dx
2
FOC 2bx=0
⇒x*=0
d y
= 2b
2
dx
f ′′ x * = 2b
y = a+bx+cx2+ex3
dy
= b + 2cx + 3ex 2
dx
Anything raised to the 0
power is 1!
2
d y
= 2c + 6ex
2
dx
Review of the FOC and SOC for an unconstrained function of one variable:
2
dy
d
y
= 0 ), then look at SOC by substituting x* into
Find x* from the FOC (
dx
dx 2
to see if second derivative evaluated at x* is &gt;, &lt;, or = 0.
• The derivative of the natural log of x is the
inverse of X,
dy d ln x 1
=
= .
dx dx x
c
dbln(x
) bc because
The actual rule is
=
dx
x
dclnx c
=
ln(x ) = clnx and
dx
x
dbclnx bc
c
=
bln(x ) = bclnx and
dx
x
May not always
dy
5
⋅
2
If y = 3 + 5ln(x 2 ),
=
, have the
dx
x
parentheses.
c
The exponential function rule dy
bx
For y = a , dx = the derivative of the exponent
bx
(bx) times the original function ( a
times
the natural log of the constant (a).
)
bx
da
dbx bx
=
a lna = ba bx lna,
dx
dx
x
da
x
= a ln a
dx
Can give constants different labels than above.
x
dy
x
If y = ac ,
then
= ac ln c.
dx
x
bx
de
x
x
x
de
bx
bx
bx
= e ln e = e (1) = e
= be ln e = be (1) = be ,
dx
dx
e is the base of natural logarithms = 2.71828, so that lne = 1.
The Sum of Two Functions Rule –
d[f ( x ) + g( x )]
= f ′( x ) + g′( x ).
dx
The derivative of the sum of two functions
of x is the sum of derivatives of the two
functions:
Example: If y = ax +
bx2,
dy
then
= a +2bx.
dx
The Product Rule – The derivative of
the product of two functions of x is the first
times the derivative of the second plus the
second times the derivative of the first.
d[f ( x ) ⋅ g( x )]
= f ( x )g′( x ) + g( x )f ′( x )
dx
Example: y = (2x-1)(2-9X2)
dy
2
= (2X − 1)(−18X) + (2 − 9 x )(2)
dx
= (−36 x 2 + 18x ) + (4 − 18x 2 )
= 4 + 18x − 54 x 2
The Quotient Rule - The derivative of
the quotient of two functions of x is the derivative of the
top times the bottom minus the top times the derivative
of the bottom all divided by the bottom squared.
 f (x) 

d
 g ( x )  = f ′( x )g ( x ) − f ( x )g′( x )
dx
[g( x )]2
Ex 1: If
Ex 2: If
Or
(
)
2
2
3
4
2
x3
+
−
+
dy
3
x
x
2
x
(
2
x
)
x
6
x
y= 2
,
=
= 4
4
2
x + 2 dx
x + 4x + 4
x + 4x 2 + 4
( )
dy 0 x 4 − 1( 4 x 3 ) − 4 x 3
−4
1
−5
y= 4 ,
4
x
−
=
=
=
=
x
dx
x8
x8
x5
y=
1
= x −4 ,
4
x
dy
= −4 x −5
dx
The Chain Rule - Allows
differentiation of complex functions in
terms of elementary functions.
If y = f ( x )
and x = g ( z ) so that
y = f (g (z)), then
dy dy dx
df
dg df dg
=
⋅
=
⋅
=
⋅
dz dx dz dg (z) dz dx dz
Examples:
1
3
y
=
(
z
+
2
z
)
, x = z + 2z, so y = x
1.
2
2
1
3
− 2  dx
dy  dy 1 2

3
=  = (z + 2z)  = 2z + 2 
dz  dx 3

 dz
−2
2(z + 1)
1 2
3
= z + 2 z ( 2 z + 2) =
2
2
3
3(z + 2z) 3
(
)
Examples:
w
w
dy
= 2(3x + 1)3 = 18x + 6
2. y = (3x + 1) ,
dx
2
w
3.
y = ln(x − 4x + 2)
2
dw
dx
dy
1
2x − 4
= 2
⋅ (2x − 4) = 2
x − 4x + 2
dx x − 4x + 2
dy
dw
Functions of More than One Variable
“Partial” derivatives – May want to know how y
changes as one of several x variables changes,
ceteris paribus (as in demand curve).
y = f (x 1 , x 2 , x 3 ,..., x n )
Several derivatives can be taken; one with respect to
each Xi holding the other variables constant.
Denoted as: ∂y
= f xi = fi
∂x i
When taking this derivative, all other xi’s are treated
as constants. Fortunately, the rules for derivatives
of functions of one variable apply to partial
derivatives also. We simply treat the other x’s as
constants.
Second Order Partial Derivatives are analogous to
the early one-variable case. Denoted as:
“cross second partial” i≠j
∂2y
∂x i ∂x j
= f ij
“own second partial”
i=j
There are four combinations in the two variable case, fij, fii, fji, and fjj; but fij
and fji are equal (Young’s Theorem). The order of differentiation does not
matter.
Partial derivatives reflect the units of measurement for y and x. If y =
quantity of apples demanded per year in 100 pound units and x is the price
of apples per pound, ∂y/∂x measures the change in the quantity of apples
demanded in 100 pound units per year for a dollar change in the apple price,
ceteris paribus. If demand were measured in pounds per year, the partial
derivative would be 100 time larger than when demand is measured in 100
pounds per year.
Example 1: Q = 1000 − 20P 2 + 36I 2
∂ 2Q
= −40
∂P∂P
∂ Q
∂Q
= 0,
= −40P,
∂P∂I
∂P
2
∂Q
∂
Q
= 72I,
= 0,
∂I
∂I∂P
2
∂ 2Q
= 72
∂I∂I
Example 2: Q = 1000 − 20P + 36I + IP
2
∂Q
= −40P + I,
∂P
∂ Q
= 1,
∂P∂I
∂Q
= 72I + P,
∂I
∂ 2Q
= 1,
∂I∂P
2
2
∂ 2Q
= −40
∂P∂P
∂ 2Q
= 72
∂I∂I
Total Differential
The total differential shows what happens to y
as both x1 and x2 change simultaneously.
The total differential is dy = ∂f dx1 + ∂f dx 2
∂x1
∂x 2
To find a maximum or minimum for y (where dy
= 0 as x1 and x2 change), the FOC is that all ∂f ∂x = 0
because then dy = 0! That is, all the first
partial derivatives must be 0. This happens at
f2
the “top of the Hill” in x2
three dimensions or
f1
the “bottom of the hole”.
i
x1
Example 1: To find max. or min. for y = − x1 + 2x1 − x 2 + 4x 2 + 5,
2
2
we must find where both first partials equal zero (FOCs) and
solve the FOCs simultaneously for x1* and x *2 .
Plug these
∂y
*
2 x1 = 2
x1 = 1 optimal values
= −2 x1 + 2 = 0
FOC 1:
∂x1
into y to get
y
∂
FOC 2:
2x 2 = 4
x *2 = 2 y* = 10.
= −2 x 2 + 4 = 0
∂x 2
Example 2: Optimize y = − x1 + 2x1 − x 2 + 4x 2 + 5 − x1x 2 ,
FOC 1: ∂y = −2x1 + 2 − x 2 = 0 ⇒ x 2 = −2x1 + 2
∂x1
FOC 2: ∂y = −2x + 4 − x = 0 ⇒ -2(-2x + 2) + 4 − x = 0
2
1
1
1
∂x 2
⇒ x1* = 0 and x *2 = 2 Plug these optimal values into y to get y* = 9.
2
2
These FOCs are not sufficient; could be a maximum, a minimum, or a saddle
point. All we know is that it is a flat spot! Need SOCs (later).
Implicit Functions
• Usually express functions explicitly as
y = f(x1, x2 ,…, xn) where y is a function of xi.
• However, some functions will be expressed
implicitly as 0 = f(x1, x2 ,…, xn, y) where
there is no dependent variable.
• Can find:
dy
fi
=−
dx i
fy
dx i
fy
=−
and
dy
fi
or
Remember the negative sign.
fj
dx i
=−
dx j
fi
Example
f(x, y) = x + y − 2x + 4y + 4 = 0
2
f x = 2x − 2
2
f y = 2y + 4
dy
2x − 2
x −1
1− x
=−
=−
=
dx
2y + 4
y+2
y+2
dx
2y + 4
y+2
Remember the sign!
=−
=
dy
2x − 2 1 − x
Examples of implicit functions are production possibility
curves, indifference curves, budget constraints.
Envelope Theorem
• Deals with how the optimal value of the
function (y*) changes as the value of
one of the parameters changes (What
are parameters?)
dy
• Theorem states that one can calculate da
by holding independent variables at
their optimal values in the function and
∗
∂
y
∗
*
*
*
finding
y
=
f(x
,
x
,...,
x
.
1
2
n)
∂a
Example: y = − x + ax + bx
2
dy
= −2x + a + b = 0,
First, find the FOC:
dx
a+b
*
Second, solve FOC for x:
x =
2
Third,
2
2
substitute x*
+
a
b
(
a
b
)
a
b
a
b
+
+
+

 



∗
into y to get: y = −
=
 + b
 + a
 2 
 2 
∗
 2 
Forth, partially differentiate y with respect to a:
(Partial derivative because b is constant)
∗
4
∂y a + b
=
2
∂a
Example cont’d
∗
From previous slide
∗
Thus,
∂y
∂a
∂y a + b
=
∂a
2
depends only on values of a and b, but no variables!

∂y*
=3
a = 4,

Assume b=2 then:
∂a


∂y*
∂y*
a+2
=
= a = 2,
= 2
∂a
∂a
2


∂y*
=1
a = 0,
∂a

In many-variable case, solve all FOC simultaneously for xi*(a) *
and substitute into y to get y* = f(x1*, x2*, …, xn*), then dy = ∂y .
da
∂a
Constrained Optimization
• Because we are looking at scarce
resources, we usually must find an
optimum value subject to one or more
constraints (limits). These may prevent
us from achieving the maximum (or
minimum) of the objective function.
Lagrangian Multiplier Method
Mathematical “voodoo” to short-cut optimization subject to a
constraint. The following is an important format for setting up a
constrained optimization problem.
Max (Min):y = f ( x1 , x 2 , x 3 ,..., x n ) Objective function to be optimized
s.t. 0 = g( x1 , x 2 , x 3 ,..., x n ) Constraint
Implicit form!
Set up L = f ( x1 , x 2 , x 3 ,..., x n ) + λg( x1 , x 2 , x 3 ,..., x n )
Where λ is the Lagrangian multiplier.
Constraint example: x1 + 2x2 + x3 = 3
0 = 3 − x1 − 2x 2 − x 3 = g( x1 , x 2 , x 3 ,..., x n )
Value of
constraint
Find FOC for each xi and λ:
∂L
= f1 + λ g 1 = 0
∂x 1



∂L

= f 2 + λg 2 = 0

∂x 2


∂L
= f 3 + λg 3 = 0

∂x 3





∂L

= f n + λg n = 0

∂x n


∂L
= g ( x1 , x 2 ,..., x n ) = 0
∂λ



FOC
Lagrangian Multiplier Method
(cont)
Solve these n + 1 FOCs simultaneously to find
* * *
*
*
values for x1 , x 2 , x 3 ,..., x n , and λ .
These are optimum values of the xis given the
constraint.
*
i
*
Substituting the x into y will give y .
These conditions are necessary, but not
sufficient for an optimum. Look at Second
Order Conditions (SOC) later.
You can use the Lagrangian method for any
number of xi (i=1…n)and for any number of
constraints. Just add another λj (j=1…m) for
each constraint, gj (j=1…m).
L = f(x1 ,..., x n ) + λ1g1(x1 ,..., x n ) + λ2g2(x1 ,..., x n )
Interpretation of λ
Solve the first FOC for λ to get:
f1
f1 + λg1 = 0, λg1 = −f1 , λ =
− g1
fi
Do it for all n FOC to get: λ =
− gi
Because all ratios equal λ, all ratios are equal.
f1
f2
fn
λ=
=
= ... =
− g1 − g 2
− gn
• At optimum (Max or Min of y), the negative of the
ratios of the partial derivatives (objective/constraint)
with respect to each xi are equal and equal to λ.
• The fi shows the marginal contribution (benefit) of xi
to y (the objective function), while the –gi shows the
marginal cost of xi in the constraint. That is, –gi
shows how much the other xis must be reduced
when xi is increased slightly, and therefore, how
much y is foregone if the constraint is to hold.
• Because λ = the ratio of marginal benefit
to marginal cost for all xi, all these ratios
are equal. This condition is necessary
for a maximum (or minimum). That is,
the Margin Benefit/Marginal Cost ratios
must be equal for all xi and equal to λ. If
one xi ratio were larger than the others,
then more xi should be used and less of
other xis should be used until the ratios
are equal again.
• λ* indicates how y * would be affected by relaxing the
constraint slightly. λ* is the shadow price for a unit of
the resource represented by the constraint (the value
of an additional unit of the resource). If λ* = 0, the
constraint is redundant and the optimum is
unconstrained. An additional unit of the resource
would be worth nothing in regard to increasing y.
• If λ* is large, relaxing the constraint would be very
beneficial because the benefit/cost ratio is high.
• Relaxing the constraint refers to increasing
(decreasing) the value of the constraint in a
maximization (minimization) problem!
Duality
• Any constrained Max (Min) problem
(called the primal problem) is
associated with a constrained Min (Max)
problem (called the dual problem).
– Primal (Dual) may be: Max Utility s.t.
Income; or Min Cost s.t. Output level.
– Dual (Primal) would be: Min Expenditures
s.t. Utility level; or Max Output s.t. Cost
level.
Always two ways to look at any constrained
optimization problem.
Example: Primal
Min: f ( x, y) = z = x 2 − y 2 + 3xy + 5x
x − 2y = 0
s.t.
L = x − y + 3xy + 5x + λ(0 − x + 2y)
2
1.
2
∂L
= 2 x + 3y + 5 − λ = 0
∂x
2.
∂L
= −2 y + 3x + 2λ = 0
∂y
3.
∂L
= −x + 2y = 0
∂λ
FOCs
Primal example (cont.) Simultaneous
Solution
1. Multiply FOC 1 by 2 and add it to FOC 2. 4x + 6y + 10 − 2λ = 0
3x − 2y + 0 + 2λ = 0
2. Solve FOC 3 for x to get x = 2 y.
7x + 4y + 10 = 0
3. Substitute x into above result to get: 7(2 y) + 4 y + 10 = 0.
This gives:
y* = − 5 9 .
4. Substitute y* into x=2y from FOC 3 to get: x * = 2( − 5 9) = − 10 9 .
5. Substitute x * and y*into FOC 1 to get: λ* = 10 9 .
6. Substitute x * and y*into z to get: z * = − 225 81.
*
*
*
7. Optimal solution is: x * = − 10 9 , y = − 5 9 , z = − 225 81, λ = 10 9 .
If you could relax the constraint by one unit (decrease it by one unit
from 0 to -1), λ* shows that z*would decrease by 10/9 of a unit.
Example: Dual
Max:
w = x − 2y
s.t.:
x 2 − y 2 + 3xy + 5x = − 225 81
From the
constraint of the
primal.
From the
objective
function of the
primal assuming
optimality.
The constraint in implicit form is: 0 = –225/81 – f(x,y).
L = x − 2y + λ (− 225 81 − x + y − 3xy − 5x)
D
2
2
Dual example (cont.)
FOCs
∂L
= 1 − 2λ D x − 3λ D y − 5λ D = 0
∂x
∂L
= −2 + 2λ D y − 3λ D x = 0
∂y
∂L
225 − x 2 + y 2 − 3xy − 5x = 0
=
−
81
∂λ D
Solve these FOCs simultaneously to get:
x * = − 10 9 ,
y* = − 5 9 ,
λ D* = 9 10 , w * = 0
If we could relax (increase) the constraint by 1 unit, the optimal
value of the objective function ( w * ) would increase by 9/10 of a
unit. Notice that values for x* and y* are common to the primal
and dual, while the value of λD* for the dual is the reciprocal of
λ* for the primal.
Envelope Theorem in Constrained Max.
• Theorem can be used as discussed earlier to
examine how a function’s optimum value (y*)
changes as a parameter in the problem
changes. In the case at hand, differentiate
optimal Lagrangian expression with respect
to the parameter:
Max: y = f ( x1 , x 2 ; a )
st: 0 = g( x1 , x 2 ; a )
L = f ( x 1 , x 2 ; a ) + λg ( x 1 , x 2 ; a )
Solve FOC for x1*, x2* and substitute into L to get L*,
which is a function of parameters only.
Envelope Theorem …(cont.)
dy ∂L
=
da ∂a
*
*
Only difference is that y* is a
constrained optimum.
dy* * ∂L*
If a is the value of the constraint, then
=λ =
da
∂a
is the shadow price of the resource in the constraint.
See the section in the text on Inequality Constraints (pp. 4547) for the Lagranian methods for solving problems with
inequality constraints.
Warnings:
Not all optimization problems can be solved with calculus.
Functions must be continuous at optimum. If not, use Linear
Programming methods.
Further discussion of FOC and SOC
SOC for functions of one independent
variable and no constraints: y = f(x).
y
′
f
=
0

FOC
Max
SOC f ′′ x * &lt; 0}concave
y
FOC f ′ = 0

Min

SOC f ′′ x * &gt; 0}convex

y*
concave
x
X*
convex
y*
X*
x
Suppose
y = f ( x1 , x 2 )
Two independent variables, no constraint.
FOC
f1 = 0  Indicates a flat spot—

f 2 = 0 max, min, saddle point.
Logic suggests that the own second partials
*
*
evaluated at x1 and x 2 would indicate which
(max or Min), but….
f11 x * , x * &lt; 0  SOC
1
2

 Max
f 22 x * , x * &lt; 0
1
2

f11 x * , x * &gt; 0 
1
2
 Min

f 22 x * , x * &gt; 0
1
2

Assumes only x1 or x2 is
changing!
However, what if both x1 and x2 are
changing? Must consider a third SOC
condition (partial derivatives are evaluated at
X1* and X2* for SOCs):
2
11 22
12
f f −f &gt; 0
This condition must hold when evaluated at the
optimal values of x1 and x2. This condition is one
of three sufficient conditions (SOCs) when f is a
function of two variables (unconstrained).
Summary of FOC and SOC for functions of
two independent variables and no constraint.
y = f(x1 , x 2 )
FOC for maximum or minimum
f1 = 0 
f 2 = 0
SOC for a maximum:
f11 &lt; 0, f 22 &lt; 0

2
f11f 22 − f12 &gt; 0
Concave function
SOC for a minimum:
f11 &gt; 0, f 22 &gt; 0
 Convex function
2
f11f 22 − f12 &gt; 0 
• f1 and f2 locate a flat spot.
• f11 and f22 distinguish between a minimum
and a maximum.
• Think of a saddle in case of a minimum. The
flat spot on the saddle is not the lowest point
• f11f22 - f122 &gt; 0 confirms that it is not a saddle
point.
Constrained Optimization
FOC
The SOC for constrained maximization of a function of two
variables, subject to a linear constraint (or its dual) is:
f11f 22 − 2f12 f1f 2 + f 22 f12 &lt; 0, evaluated at x1*and x *2 .
∂L

=0
If a constrained function has this property, it is strictly quasi-concave.
∂x1
In economic theory, we usually use this problem (e.g., Max. a utility
∂L
= 0 function, s.t. a linear budget constraint (primal), or Min. linear
∂x 2
expenditures, s.t. a fixed level of the utility function (dual).
…
The SOC for constrained minimization of a function of two
∂L
= 0 variables, subject to a linear constraint (or its dual) is:
∂x n
f11f 22 − 2f12 f1f 2 + f 22 f12 &gt; 0, evaluated at x1* and x *2 .

∂L
= 0 If a constrained function has this property, it is strictly quasi-convex.
∂λ 1
Solve the FOC simultaneously to get the optimal solution and plug
…
∂L
=0
∂λ m
the xi* into the appropriate expression above (SOC) to see if it is
satisfied (or just determine the signs of the partial derivatives and
determine if the expression has the appropriate sign).
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