partial differentiation & application

Partial Differentiation & Application

Week 9

1. Function with two variables

2. First Partial Derivatives

3. Applications of First Partial Derivatives Cob-Douglas Production Function Substitute and Complementary Commodities

4. Second Partial Derivatives

5. Application of Second Partial Derivatives Maxima and Minima of Functions of Several Variables* Lagrange Multipliers*

*Additional topic

Contents:

Functions of Two Variables

A Function of Two Variables

A real-valued function of two variables, f, consists of:

1. A set A of ordered pairs of real numbers (x, y) called the domain of the function.

2. A rule that associates with each ordered pair in the domain of f one and only one real number, denoted by z = f (x, y).

Function of Two Variables

Independent variablesDependent variable

a. A company produces two products, A and B. The joint cost function (in RM) is given by:

b. Country workshop manufactures both furnished and unfurnished furniture for home. The estimated quantities demanded each week of its desks in the finished and unfinished version are x and y units when the corresponding unit prices are

respectively.

6000857507.0),( 2 yxxyxfC

Partial Derivatives: Application

Example

yxq

yxp

25.02.0160

1.02.0200

Examples of problems with two variables.

Let f be the function defined by2 3( , ) 3 2 .f x y x y y

Find (0,3) and (2, 1).f f

2 3(0,3) 3 0 (3) 2 3f

25

2 3(2, 1) 3 2 ( 1) 2 1f

15

Example


Let f be the function defined by

Find

Example

3),(

yx

xyxf

)2,3(),3,2( ff

23

3233

)2,3(

1332

2)3,2(

f

f


Find the domain of each function2a. ( , ) 3 2f x y x y

Since f (x, y) is defined for all real values of x and y (x and y is linear function), the domain of f is the set of all points (x, y) in the xy – plane.

b. g( , )2 3

xx y

x y

g(x, y) is defined as long as 2x + y – 3 is not 0. So the domain is the set of all points (x, y) in the xy – plane except those on the line y =–2x + 3.

Example


Let f be the function defined by

Find the domain of the function

Question

g(x, y) is defined as long as

So the domain is the set of all points (x, y) in the xy – plane except those on the line y=25

2

5.02

y2.05

yx03.01000100)y,x(f

02.05 2 y

25

02.05

02.05 2

y

y

y


10

Acrosonic manufactures a bookshelf loudspeaker system that may be bought fully assemble or in a kit. The demand equations that relate the unit prices, p and q to the quantities demanded weekly, x and y, of the assembled and kit versions of the loudspeaker systems are given by

What is the weekly total revenue function R (x,y)?

1 1 1 3300 240

4 8 8 8p x y q x y

2 2

( , )

1 1 1 3300 240

4 8 8 8

1 3 1300 240

4 8 4

R x y xp yq

x x y y x y

x y xy x y

Example


Recall the Graph of Two Variables

Ex. Plot (4, 2)

(4, 2)

Ex. Plot (-2, 1)

Ex. Plot (2, -3)

(2, -3)(-2, 1)


Graphs of Functions of Two Variables

Three-dimensional coordinate system: (x, y, z)

Ex. Plot (2, 5, 4)

z

y

x

24

5


Graphs of Functions of Two Variables

Ex. Graph of f (x, y)= 4 – x2 – y2


First Partial Derivatives of f (x, y).

f (x, y) is a function of two variables. The first partial derivative of f with respect to x at a point (x, y) is

0

( , ) ( , )limh

f f x h y f x y

x h

provided the limit exits.

Partial Derivatives

),(),( yxfx

yxff xx

First Partial Derivatives of f (x, y).

f (x, y) is a function of two variables. The first partial derivative of f with respect to y at a point (x, y) is

provided the limit exits.

0

( , ) ( , )limk

f f x y k f x y

y k

Partial Derivatives

),(),( yxfy

yxff yy

To get partial derivatives…. To get assume y is a constant and

differentiate with respect to x

To get assume x is a constant and

differentiate with respect to y

xf

yf

Partial Derivatives

yxxyyxf 22),(

xyyyxyyxf x 2)2()1(),( 22

22 2)1()2(),( xxyxyxyxf y yxxyyxf 22),(

Example

Example

2( , ) 3 lnf x y x y x y

6 lnxf xy y 2 13yf x x

y

2

( , ) xy yg x y e

2

2 1 xy yyg xy e

Compute the first partial derivativesExample

Example Compute the first partial derivatives

yxyx eyg

22

Partial Derivatives

Compute the first partial derivativesExample

Example Compute the first partial derivatives

32 yx ff

2 3( , ) 3 2 .f x y x y y

432),( yxyxf

22 336 yxfxyf yx

Partial Derivatives

19

The Cobb-Douglas Production Function

• a and b are positive constants with 0 < b < 1.

• x stands for the money spent on labor, y stands for the cost of capital equipment.

• f measures the output of the finished product and is called the production function

1( , ) b bf x y ax y

fx is the marginal productivity of labor.fy is the marginal productivity of capital.

Partial Derivatives: Application of First Partial Derivatives

20

A certain production function is given by

units, when x units of labor and y units of capital are used. Find the marginal productivity of capital when labor = 81 units and capital = 256 units.

1/ 4 3/ 4( , ) 28f x y x y

1/ 4

1/ 4 1/ 421 21y

xf x y

y

1/ 481 3

21 21 15.75 256 4

So 15.75 units per unit increase in capital expenditure.

Example

yfWhen labor = 81 units and capital = 256 units,


21

A certain production function is given by

units, when x units of labor and y units of capital are used. Find the marginal productivity of labor when labor = 81 units and capital = 256 units.

1/ 4 3/ 4( , ) 28f x y x y

So 49.78 units per unit increase in labor expenditure.

Question

When labor = 81 units and capital = 256 units,

4/34/34/34/3 74

128 yxyxf x

78.49)256()81(7 4/34/3 xf


22

Substitute and Complementary Commodities

Suppose the demand equations that relate the quantities demanded, x and y, to the unit prices, p and q, of two commodities, A and B, are given by

x = f(p,q) and y = g(p,q)


23

Substitute and Complementary Commodities

Two commodities A and B are substitute commodities if

0 and 0f g

q p

0 and 0f g

q p

Two commodities A and B are complementary commodities if


24

The demand function for two related commodities are x = ae q-p

y = be p-q

The marginal demand functions are x = - ae q-p y = be p-q

p p

x = ae q-p y = - be p-q

q q Because x/q > 0 and y/p > 0, the two commodities are substitute commodities.

Example


25

In a survey it was determined that the demand equation for VCRs is given by

The demand equation for blank VCR tapes is given by

Where p and q denote the unit prices, respectively, and x and y denote the number of VCRs and the number of blank VCR tapes demanded each week. Determine whether these two products are substitute, complementary, or neither.

0.5( , ) 10,000 10 qx f p q p e

( , ) 50,000 4000 10y g p q q p

Question


Because x/q < 0 and y/p < 0, the two commodities are complementary commodities.

qeq

x

p

x

5.05.0

10

0.5( , ) 10,000 10 qx f p q p e ( , ) 50,000 4000 10y g p q q p

4000

10

q

y

p

y


27

2

2xx x

ff f

x x

Second-Order Partial Derivatives

2

2yy y

ff f

y y

2

xy x

ff f

y x y

2

yx y

ff f

x y x

Partial Derivatives: Second-Order Partial Derivatives

2( , ) 3 lnf x y x y x y 6 lnxf xy y 2 1

3yf x xy

Find the second-order partial derivatives of the functionExample

Example Find the second-order partial derivatives of the function

yxf

yxf

y

xfyf yxxyyyxx

16

166 2

yff yx 43

2a. ( , ) 3 2f x y x y

0f 0f 4f 0f yxxyyyxx


29

Find the second-order partial derivatives of the function2

( , ) xyf x y e

2 22xy xyxf e y e

x

2 2

2xy xyyf e xye

y

24 xy

xxf y e

Example

2 22 1xyyxf ye xy


2xyxy2xyxy xy1ye2xye2yye2f

222

2xyxyxyyy xy21xe2xy2eyex2f

222

30

Maximum and Minimum of Functions of Several

Variables

31

Relative Extrema of a Function of Two VariablesLet f be a function defined on a region R containing (a, b).

f (a, b) is a relative maximum of f if ( , ) ( , )f x y f a b

( , ) ( , )f x y f a bf (a, b) is a relative minimum of f if

for all (x, y) sufficiently close to (a, b).

for all (x, y) sufficiently close to (a, b).

*If the inequalities hold for all (x, y) in the domain of f then the points are absolute extrema.

Partial Derivatives: Application of Second Partial Derivatives

32

Critical Point of f

A critical point of f is a point (a, b) in the domain of f such that both

, 0 and , 0f f

a b a bx y

or at least one of the partial derivatives does not exist.

Partial Derivatives: Application of Second Partial Derivatives

33

Determining Relative Extrema1. Find all the critical points by solving the system

0, 0x yf f

2. The 2nd Derivative Test: Compute2( , ) xx yy xyD x y f f f

( , )D a b ( , )xxf a b Interpretation

+

+

+

–

0

– Relative min. at (a, b)

Relative max. at (a, b)

Test is inconclusive

Neither max. nor min. at (a, b) saddle point

34

Ex. Determine the relative extrema of the function2 2( , ) 2f x y x x y

2 2 0 2 0x yf x f y So the only critical point is (1, 0).

2(1,0) 2 2 0 4 0D 1,0 2 0xxf and

2, 0xx yy xyf f f

So f (1,0) = 1 is a relative maximum

35

ApplicationEx: The total weekly revenue (in dollars) that Acrosonic realizes in producing and selling its bookshelf loudspeaker systems is given by

where x denotes the number of fully assembled units and y denotes the number of kits produced and sold each week. The total weekly cost is given by

Determine how many assembled units and how many kits Acrosonic should produce per week to maximize its profit.

2 21 3 1, 300 240

4 8 4R x y x y xy x y

, 180 140 5000C x y x y

36

2 2

, , ,

1 3 1120 100 5000

4 8 4

P x y R x y C x y

x y xy x y

1 1120 0 1

2 43 1

100 0 24 4

x

y

P x y

P y x

37

2 480y x

3 12 480 100 0

4 46 1440 400 0

208

yP x x

x x

x

Substitute in 1

Substitute in 2

64y Substitute this value into the equation 2 480y x

Therefore, P has the critical point (208,64)

38

1 1 3

2 4 4xx xy yyP P P

2

1 3 1 5,

2 4 4 16D x y

Since, and , the point (208,64) is a relative maximum of P.

208,64 0D 208,64 0xxP

39

Lagrange Multipliers

Reading: Mizrahi and Sullivan, 8th ed., 2004, WileyChapter:17.5

40

Method of Lagrange Multipliers

Suppose that, subject to the constraint g(x,y)=0, the function z=f(x,y) has a local maximum or a local minimum at the point

.

Form the function

0 0( , )x y

, , , ,F x y f x y g x y

A method to find the local minimum and maximum of a function with two variables subject to conditions or constraints on the variables involved.

41

0 1

0 2

, 0 3

F f g

x x xF f g

y y y

Fg x y

Then there is a value of such that is a solution of the system of equations

0 0( , , )x y

provided all the partial derivatives exists.

42

Steps for Using the Method of Lagrange Multipliers

Step 1: Write the function to be maximized (or minimized) and the constraint in the form:

Find the maximum (or minimum) value of

subject to the constraint

Step 2: Construct the function F:

,z f x y

, 0g x y

, , , ,F x y f x y g x y

43

Step 3: Set up the system of equations

Step 4: Solve the system of equations for x, y and . Step 5: Test the solution to determine

maximum or minimum point.

0 1

0 2

, 0 3

F

xF

y

Fg x y

0 0( , , )x y

44

Find D* = Fxx . Fyy - (Fxy)2

If D* 0 Fxx 0 maximum point

Fxx 0 minimum point

D* 0 Test is inconclusive

Step 6: Evaluate at each solution found in Step 5.

,z f x y 0 0( , , )x y

45

Example:Find the minimum of

f(x,y) = 5x2 + 6y2 - xysubject to the constraint

x+2y = 24

Solution:F(x,y, ) = 5x2 + 6y2 - xy + (x + 2y - 24)

Fx = F = 10x - y + ; Fxx = 10x

Fy = F = 12y - x + 2 ; Fyy = 12y

F = F = x + 2y - 24 ; Fxy = -1

46

The critical point,10x - y + = 012y - x + 2= 0x + 2y - 24= 0

The solution of the system is x = 6, y = 9, = -51

D*=(10)(12)-(-1)2=119>0

Fxx = 10>0

We find that f(x,y) has a local minimum at (6,9).

f(x,y) = 5(6)2+6(9)2-6(9)= 720

47

Example

A manufacturer produces two types of engines, x units of type I and y units of type II. The joint profit function is given by

to maximize profit, how many engines of each type should be produced if there must be a total of 42 engines produced?

2, 3 6P x y x xy y

48

2, 3 6z P x y x xy y Maximize

Subject to constraint ( , ) 42 0g x y x y

2

, , , ,

3 6 42

F x y P x y g x y

x xy y x y

2 3 0 1 ; 2

3 6 0 2 ; 0

42 0 3 ; 3

xx

yy

xy

Fx y F

xF

x Fy

Fx y F

The solution of the system is 33 9 93.x y

49

2 0xxF

2* (2)(0) (3) 9 0D

The test in inconclusive.

partial differentiation & application

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y f y x

y f x x

x y xy

x ln y

ln y f y

y isf f x

xy f y

line y