Integrations with applications for engineering students

Models with Multiple
Explanatory Variables
Chapter 2 assumed that the dependent variable (Y) is affected by only ONE
explanatory variable (X).
Sometimes this is the case.
Example: Age = Days Alive/365.25
Usually, this is not the case.
Example: midterm mark depends on:
 how much you study
 how well you study
 intelligence, etc
1

4. Multi Variable Examples:
Demand = f( price of good, price of substitutes, income, price of compliments)
Consumption = f( income, tastes, wages)
Graduation rates = f( tuition, school quality, student quality)
Christmas present satisfaction = f (cost, timing, knowledge of person, presence of
card, age, etc.)
2

4. The Partial Derivative
It is often impossible analyze ONE variable’s impact if ALL variables are changing.
Instead, we analyze one variable’s impact, assuming ALL OTHER VARIABLES
REMAIN CONSTANT
We do this through the partial derivative.
This chapter uses the partial derivative to expand the topics introduced in chapter
2.
3

4. Calculus and Applications involving More
than One Variable
4.1 Derivatives of Functions of More Than One
Variable
4.2 Applications Using Partial Derivatives
4.3 Partial and Total Derivatives
4.4 Unconstrained Optimization
4.5 Constrained Optimization
4

4.1 Partial Derivatives
Consider the function z=f(x,y). As this function takes into account 3 variables, it
must be graphed on a 3-dimensional graph.
A partial derivative calculates the slope of a
2-dimensional “slice” of this 3-dimensional graph.
The partial derivative ∂z/∂x asks how x affects z while y is held constant (ceteris
paribus).
5

In taking the partial deriv
ative, all other variables are kept constant and hence treated as constants (the
derivative of a constant is 0).
There are a variety of ways to indicate the partial derivative:
1) ∂y/∂x
2) ∂f(x,z)/∂x
3) fx(x,z)
Note: dy=dx is equivalent to ∂y/∂x if y=f(x); ie: if y only has x as an explanatory
variable.
(Therefore often these are used interchangeably in economic shorthand)
6

Let y = 2x2
+3xz+8z2
∂y/ ∂x = 4x+3z+0
∂y/ ∂z = 0+3x+16z
(0’s are dropped)
Let y = xln(zx)
∂ y/ ∂ x = ln(zx) + zx/zx
= ln(zx) + 1
∂ y/ ∂ z = x(1/zx)x
=x/z
7

Let y = 3x2
z+xz3
-3z/x2
∂ y/ ∂ z=3x2
+3xz2
-3/x2
∂ y/ ∂ x=6xz+z3
+6z/x3
Try these:
z=ln(2y+x3
)
Expenses=sin(a2
-ab)+cos(b2
-ab)
8

4.1.1 Higher Partial Derivatives
Higher order partial derivates are evaluated exactly like normal higher order
derivatives.
It is important, however, to note what variable to differentiate with respect to:
From before:
Let y = 3x2
z+xz3
-3z/x2
∂ y/ ∂ z=3x2
+3xz2
-3/x2
∂ 2
y/ ∂ z2
=6xz
∂ 2
y/ ∂ z ∂ x=6x+3z2
+6/x3
9

4.1.1 Young’s Theorem
From before:
Let y = 3x2
z+xz3
-3z/x2
∂ y/ ∂ x=6xz+z3
+6z/x3
∂ 2
y/ ∂ x2
=6z-18z/x4
∂ 2
y/ ∂ x ∂ z=6x+3z2
+6/x3
Notice that ∂2
y/∂x∂z=∂2
y/∂z∂x
This is reflected by YOUNG’S THEOREM: order of differentiation doesn’t matter
for higher order partial derivatives
10

4.2 Applications using Partial Derivatives
As many real-world situations involve many variables, Partial Derivatives can be
used to analyze our world, using tools including:
 Interpreting coefficients
 Marginal Products
11

4.2.1 Interpreting Coefficients
Given a function a=f(b,c,d), the dependent variable a is determined by a variety of
explanatory variables b, c, and d.
If all dependent variables change at once, it is hard to determine if one dependent
variables has a positive or negative effect on a.
A partial derivative, such as ∂ a/ ∂ c, asks how one explanatory variable (c), affects
the dependent variable, a, HOLDING ALL OTHER DEPENDENT VARIABLES
CONSTANT (ceteris paribus)
12

4.2.1 Interpreting Coefficients
A second derivative with respect to the same variable discusses curvature.
A second cross partial derivative asks how the impact of one explanatory variable
changes as another explanatory variable changes.
Ie: If Happiness = f(food, tv),
∂ 2
h/ ∂ f ∂tv asks how watching more tv affects food’s effect on happiness (or how
food affects tv’s effect on happiness). For example, watching TV may not
increase happiness if someone is hungry.
13

4.2.1 Corn Example
Consider the following formula for corn production:
Corn = 500+100Rain-Rain2
+50Scare*Fertilizer
Corn = bushels of corn
Rain = centimeters of rain
Scare=number of scarecrows
Fertilizer = tonnes of fertilizer
Explain this formula
14

4.2.1 Corny Example
1) Intercept = 500
-if it doesn’t rain, there are no scarecrows and no fertilizer, the farmer will
harvest 500 bushels
2) ∂Corn/∂Rain=100-2Rain
-each additional cm of rain changes corn production by 100-2Rain
-positive impact if rain < 50 cm
-negative impact if rain > 50 cm
15

4.2.1 Corny Example
3) ∂2
Corn/∂Rain2
=-2<0, (concave)
-More rain has a DECREASING impact on the corn harvest
-More rain DECREASES rain’s impact on the corn harvest by 2
4) ∂Corn/∂Scare=50Fertilizer
-More scarecrows will increase the harvest 50 for every tonne of fertilizer
-if no fertilizer is used, scarecrows are useless
16

4.2.1 Corny Example
5) ∂ 2
Corn/∂Scare2
=0 (straight line, no curvature)
-Additional scarecrows have a CONSTANT impact on corn’s harvest
6) ∂ 2
Corn/∂Scare∂Fertilizer=50
-Additional fertilizer increases scarecrow’s impact on the corn harvest by 50
17

4.2.1 Corny Example
7) ∂Corn/∂Fertilizer=50Scare
-More fertilizer will increase the harvest 50 for every scarecrow
-if no scarecrows are used, fertilizer is useless
8) ∂ 2
Corn/∂Fertilizer2
=0, (straight line)
-Additional fertilizer has a CONSTANT impact on corn’s harvest
18

4.2.1 Corny Example
9) ∂ 2
Corn/∂Fertilizer ∂Scare =50
-Additional scarecrows increase fertilizer’s impact on the corn harvest by 50
19

4.2.2 Partial Derivatives and
Marginal Product
Consider the function Q=f(L,K,)
The partial derivative is the change in output if one input (labour or capital)
increases by one.
 The partial derivative IS the MARGINAL PRODUCT of an production function
∂ Q/ ∂ L=Marginal Product of Labour (MPL)
∂ Q/ ∂ K=Marginal Product of Capital (MPK)
20

4.2.2 Cobb-Douglas Production Function
A favorite function of economists is the Cobb-Douglas Production Function of the
form
Q=aLb
Kc
Of
Where L=labour, K=Capital, and O=Other (education, technology, government,
etc.)
This is an attractive function because if b+c+f=1, the demand function is
homogeneous of degree 1. (Doubling all inputs doubles outputs…a happy
concept)
21

4.2.2 Cobb-Douglas University
Consider a production function for university degrees:
Q=aLb
Kc
Af
Where
L=Labour (ie: professors),
K=Capital (ie: classrooms)
A=Administration
22

4.2.2 Average and Marginal Products
Finding partial derivatives:
∂ Q/ ∂ L =abLb-1
Kc
Af
=b(aLb
Kc
Af
)/L
=b(Q/L) OR
=b* average product of labour
-in other words, adding an additional professor will contribute a fraction of the
average product of each current professor
-this partial derivative gives us the MARGINAL PRODUCT of labour
23

4.2.2 Cobb-Douglas Professors
For example, if 20 professors are employed by the department, and 500
students graduate yearly, and b=0.5:
∂ Q/ ∂ L =0.5(500/20)
=12.5
Ie: Hiring another professor will graduate 12.5 more students. The marginal
product of labour is 12.5
24

4.3 Total Derivatives
Often in econometrics, one variable is influenced by a variety of other variables.
Ie: Happiness =f(sun, driving)
Ie: Productivity = f(labor, effectiveness)
Using TOTAL DERIVATIVES, we can examine how growth of one variable is caused by growth in all other variables
The following formulae will combine x’s impact on y (dy/dx) with x’s impact on y, with other variables held constant
(δy/δx)
25

Assume you are increasing the square footage of a house where
AREA = LENGTH X WIDTH
A=LW
If you increase the length,
the change in area is equal
to the increase in length
times the current width:
26
Length
Area
dL
Notice that:
δA/δL=W, (partial derivative, since width is constant)
Therefore the increase in area is equal to:
dA=(δA/δL)dL

A=LW
If you increase the width,
the change in area is equal
to the increase in width
times the current length:
27
Length
Area
dW
Notice that:
δA/δW=L, (partial derivative, since length is constant)
Therefore the increase in area is equal to:
dA=(δA/δW)dW
Next we combine the two effects:

A=LW
An increase in both length
and width has the following
impact on area:
28
Length
Area
dW
Now we have:
dA=(δA/δL)dL+(δA/δW)dW+(dW)dL
But since derivatives always deal with instantaneous slope
and small changes, (dW)dL is small and ignored, resulting in:
dA=(δA/δL)dL+(δA/δW)dW
dL

29
Length
Area
dW
dA=(δA/δL)dL+(δA/δW)dW
Effectively, we see that change in the dependent variable (A),
comes from changes in the independent variables (W and L).
In general, given the function z=f(x,y) we have:
dL
dy
y
z
dx
x
z
dy
y
y
x
f
dx
x
y
x
f
dz












)
,
(
)
,
(

The key advantage of the total derivative is it takes variable
interaction into account.
The partial derivative (δz/δx) examines the effect of x on z if
y doesn’t change. This is the DIRECT EFFECT.
However, if x affects y which then affects z, we might want
to measure this INDIRECT EFFECT.
We can modify the total derivative to do this:
4.3 Total Derivative Final Form
30
dx
dy
y
z
x
z
dx
dy
y
z
dx
dx
x
z
dx
dz
dy
y
z
dx
x
z
dy
y
y
x
f
dx
x
y
x
f
dz
























)
,
(
)
,
(

4.3 Total Derivative Extension
31
dx
dy
y
z
x
z
dx
dy
y
z
dx
dx
x
z
dx
dz
dy
y
z
dx
x
z
dy
y
y
x
f
dx
x
y
x
f
dz
























)
,
(
)
,
(
Here we see that x’s total impact on z is broken up into two
parts:
1)x’s DIRECT impact on z (through the partial derivative)
2)x’s INDIRECT impact on z (through y)
Obviously, if x and y are unrelated, (δy/δx)=0, then the total
derivative collapses to the partial derivative

4.3 Total Derivative Example
32
Assume Happiness=Candy+3(Candy)Money+Money2
h=c+3cm+m2
Furthermore, Candy=3+Money/4 (c=3+m/4)
The total derivative of happiness with regards to money:
m
c
dm
dh
m
m
c
dm
dh
dm
dc
c
h
m
h
dm
dh
75
.
2
3
25
.
0
)]
4
/
1
)(
3
1
[(
]
2
3
[














4.3 Total Derivative and Elasticity
33
Total derivatives can also give us the relationship between
elasticity and revenue that we found in Chapter 2.2.3:
demand)
of
elasticity
price
is
(where
)
1
(
)
1
(















Q
dP
dTR
dP
dQ
Q
P
Q
dP
dTR
dP
dQ
P
dP
dP
Q
dP
dTR
dQ
Q
TR
dP
P
TR
dTR
PQ
TR

4.4 Unconstrained Optimization
Unconstrained optimization falls into two categories:
1) Optimization using one variable (ie: changing wage to
increase productivity, working conditions are constant)
2) Optimization using two (or more) variables (ie: changing
wage and working conditions to maximize productivity)
34

4.4 Simple Unconstrained Optimization
For a multivariable case where only one variable is
controlled, optimization steps are easy:
Consider the function z=f(x)
1) FOC:
Determine where δz/δx=0 (necessary condition)
2) SOC:
δ2
z/δx2
<0 is necessary for a maximum
δ2
z/δx2
>0 is necessary for a minimum
3) Determine max/min point
Substitute the point in (2) back into the original equation.
35

Let productivity = -wage2
+10wage(working conditions)2
P(w,c)=-w2
+10wc2
If working conditions=2, find the wage that maximizes productivity
P(w,c)=-w2
+40w
1) FOC:
δp/δw =-2w+40=0
w=20
2) SOC:
δ2
p/δw2
= -2 < 0, a maximum exists
36

P(w,c)=-w2
+10wc2
w=20 (maximum confirmed)
3) Find Maximum
P(20,4)=-202
+10(20)(2)2
P(20,4)=-400+800
P(20,4)=400
Productivity is maximized at 400 when wage is 20.
37

4.4 Complex Unconstrained Optimization
For a multivariable case where only two variable are
controlled, optimization steps are more in-depth:
Consider the function z=f(x,y)
1) FOC:
Determine where δz/δx=0 (necessary condition)
And
Determine where δz/δy=0 (necessary condition)
38

For a multivariable case where only two variable are
controlled, optimization steps harder:
Consider the function z=f(x,y)
2) SOC:
δ2
z/δx2
<0 and δ2
z/δy2
<0 are necessary for a maximum
δ2
z/δx2
>0 and δ2
z/δy2
>0 are necessary for a minimum
Plus, the cross derivatives can’t be too large compared to the
own second partial derivatives:
39
0
2
2
2
2
2
2

































y
x
z
y
z
x
z

If this third SOC requirement is not fulfilled, a SADDLE POINT
occurs, where z is a maximum with regards to one
variable but a minimum with regards to the other. (ie:
wage maximizes productivity while working conditions
minimizes it)
Vaguely, even though both variables work to increase z,
their interaction with each other outweighs this
maximizing effect
40
0
2
2
2
2
2
2

































y
x
z
y
z
x
z

Let P(w,c)=-w2
+wc-c2
+9c , maximize productivity
1) FOC:
δp/δw =-2w+c=0
2w=c
δp/δc=w-2c+9=0
w=2c-9
w=2(2w)-9
-3w=-9
w=3
2w=c
6=c
41

P(w,c)=-w2
+wc-c2
+9c
δp/δw =-2w+c=0
δp/δc=w-2c+9=0
w=3, c=6 (possible max/min)
2) SOC:
δ2
p/δw2
= -2 < 0
δ2
p/δc2
= -2 < 0, possible max
42
0
3
1
)
2
)(
2
(
2
2
2
2
2
2
2
2
2
2
2
2
2






































































c
w
p
c
p
w
p
c
w
p
c
p
w
p Maximum confirmed

P(w,c)=-w2
+wc-c2
+9c
w=3, c=6 (confirmed max)
3) Find productivity:
43
27
)
,
(
54
36
18
9
)
,
(
)
6
(
9
6
6
)
3
(
3
)
,
(
9
)
,
(
2
2
2
2
















c
w
p
c
w
p
c
w
p
c
c
wc
w
c
w
p
Productivity is maximized at 27 when wage=3 and working
conditions=6.

Typically constrained optimization consists of maximizing or minimizing an objective function with regards
to a constraint, or
Max/min z=f(x,y)
Subject to (s.t.): g(x,y)=k
Where k is a constant
44

Often economic agents are not free to make any decision they would like. They are
CONSTRAINED by factors such as income, time, intelligence, etc.
When optimizing with constraints, we have two general methods:
1) Internalizing the constraint
2) Creating a Lagrangeian function
45

4.5 Internalizing Constraints
If the constraint can be substituted into the
equation to be optimized, we are left with an
unconstrained optimization problem:
Example:
Bob works a full week, but every Saturday he has
seven hours left free, either to watch TV or read.
He faces the constrained optimization problem:
Max. Utility=7TV-TV2
+Read (U=7TV-TV2
+R)
s.t. 7=TV+Read (7=TV+R)
46

4.5 Internalizing Constraints
Max. U=7TV-TV2
+R
s.t. 7=TV+R
We can solve the constraint:
R=7-TV
And substitute into the objective function:
U=-TV2
+7TV+(7-TV)
U=-TV2
+6TV+7
47

4.5 The Lagrangian
L=z*=z(x,y)+λ(k-g(x,y))
After finding FOC’s, to confirm a maximum or minimum, the SOC is employed.
This SOC must be negative for a maximum and positive for a minimum
Note that for more terms, this function becomes exponentially complicated.
SOC’s:
48























































δxδy
z
δ
δy
δg
δx
δg
2
δx
z
δ
δy
δg
δy
z
δ
δx
δg 2
2
2
2
2
2
2
SOC

4.5 Lagrangian example
Max. U=7TV-TV2
+R
s.t. 7=TV+R
L=z*=z(x,y)+λ(k-g(x,y))
L=7TV-TV2
+R+λ(7-TV-R)
FOC:
49
2TV
-
7
0
-
2TV
-
7







TV
L
1
0
-
1







R
L
R
V
T




T
7
0
R
-
V
-
7

L

50
R
TV 



7
)
3
(
1
)
2
(
2TV
-
7
)
1
(


3
1
2
7
)
2
(
)
1
(




TV
TV
R
R
R
V





4
3
7
T
7
:
)
3
(

4.5 Lagrangian Example
51
           
0
2
0
1
1
2
2
-
1
0
1
δTVδR
z
δ
δR
δg
δTV
δg
2
δTV
z
δ
δR
δg
δR
z
δ
δTV
δg
2
2
2
2
2
2
2
2
2

























































SOC
SOC
SOC
Since the second order condition is negative, the points
found are a maximum.
Notice that we found the same answers as internalizing the
constraint.

4.5 The Lagrange Multiplier
The Lagrange Multiplier, λ, provides a measure of how much of an impact
relaxing the constraint would make, or how the objective function
changes if k of g(x,y)=k is marginally increased.
The Lagrange multiplier answers how much the maximum or minimum
changes when the constraint g(x,y)=k increases slightly to g(x,y)=k+δ
52

53
1
2(3)
-
7
2TV
-
7
4
3
2TV
-
7








R
TV
This means that if Bob gets an extra hour, his maximum utility
will increase by approximately 1.
(Alternately, if Bob loses an hour of leisure, his maximum
utility will decrease by approximately 1.)
Check:
If 8=TV+R, TV=3.5, R=4.5, U=16.75
(utility increases by approximately 1)

Constrained Optimization
So far we have discussed optimizing functions without
placing restrictions upon the values that the independent
variables can assume. Such problems are often referred to as
free maxima and minima or free optima.
However, in the real world, often
restrictions or constraints
are placed upon values of the independent
variables.
These problems are referred to as
constrained optima or constrained optimization.

constraint
Constrained
maximum
Free
maximu
m
Graphically, the difference between the free optima
and the constrained optima can be shown as:

• The free optima occurs at the peak of the
surface.
• If we specify a specific relationship between
variables and (a constraint) then the
search for an optimum is restricted to a slice
of the surface. The constrained maximum
occurs at the peak of the slice.
constraint
Constrained
maximum
Free
maximu
m

• Since economists deal with the allocation of
scarce resources among alternative uses, the
concept of constraints or restrictions is
important.
• There are two approaches to solving
constrained optima problems:
(i) substitution method
(ii) Lagrange multipliers

Substitution Method
• Consider a firm producing commodity with
the following production function:
• Without any constraints, the firm can produce
an unlimited quantity by utilizing an unlimited
amount
of and .

Substitution Method
• But suppose the firm has a budget constraint:
Let
• For simplicity, assume that the maximum
amount the firm can spend on these two inputs
is $100.
• So we have the following constraint:

Substitution Method
• Suppose the economic question facing this firm
is maximizing production subject to this budget
constraint.
• The solution via the substitution method is to
substitute:
• First, write the constraint in terms of :

Substitution Method
• Then substitute this value into the production
function, such that:
• With this substitution, the constrained
maxima problem is reduced to a free maxima
problem with one independent variable.

Substitution Method
• Now apply the usual optimization procedure:
(critical value)

Substitution Method
◦ The method of substitution is one way to
solve constrained optima problems. This is
manageable in some cases. In others, the
constraint may be very complicated and
substitution becomes complex.

Lagrange Multipliers
 The constrained optima problem can be
stated as finding the extreme value of
subject to .
 So Lagrange (a mathematician) formed the
augmented function.
denotes augmented function
will behave like the function if the constraint is foll

• Given the augmented function, the first order
condition for optimization (where the
independent variables are , and λ) is as
follows:

 Using the previous example:
note:
to be on the budget
line

• Solving these 3 equations simultaneously:

• Solving these 3 equations simultaneously
(cont’d):

• If , then
• This solution yields the same answer as the substitution
method, i.e., and .

• Economists prefer using the Lagrange technique
over the substitution method, because:
Namely for (ii), the value of λ has an
economic interpretation. (There is no
counterpart to this variable in the substitution
method).
(i) easier to handle for most cases and
(ii) provides additional information.

 Given,
where C is the level of expenditures
(budget = $100 in this example).
here the Lagrange multiplier measures the
sensitivity of to changes in the constraint
(C ).

 2nd
order conditions
Given
 1st
order conditions for extremum:

• 2nd
order conditions involve second order partial
derivatives expressed in the form of a
determinant.
• In the constraint case, we will utilize the
bordered Hessian – which is the Hessian of the
free optima case surrounded by the partial
derivatives with respect to the Lagrange
multiplier λ.

 2nd
order partials:

 2 explanatory variables and 1 constraint is the largest
size bordered Hessian we will consider for this class.
Note: partials with respect to Lagrange
multiplier (λ) form the border

 So the 2nd
order condition is:
 Let’s now return to our previous example:
◦ Recall the critical values of ,
and .

• What about the 2nd
order conditions?

Expand by 1st
row:

represents a rel max.
represents a rel max.

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