Partial Differentiation & Application

1. Partial Differentiation &
Application
Week 9

2. Contents:
1. Function with two variables
2. First Partial Derivatives
3. Applications of First Partial Derivatives
 Cob-Douglas Production Function
 Substitute and Complementary Commodities
1. Second Partial Derivatives
2. Application of Second Partial Derivatives
 Maxima and Minima of Functions of Several Variables*
 Lagrange Multipliers*
*Additional topic

3. Functions of Two Variables

4. 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).
Dependent
Independent variables variable
Function of Two Variables

5. Example Examples of problems with two variables.
a. A company produces two products, A and B.
The joint cost function (in RM) is given by:
C = f ( x, y ) = 0.07 x 2 + 75 x + 85 y + 6000
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 p = 200 − 0.2x − 0.1y
q = 160 − 0.2x − 0.25y
respectively.
Partial Derivatives: Application

6. Example
Let f be the function defined by
2 3
f ( x, y ) = 3 x y − 2 + y .
Find f (0,3) and f (2, −1).
f (0,3) = 3 ( 0 ) (3) − 2 + ( 3)
2 3
= 25
f (2, −1) = 3 ( 2 ) (−1) − 2 + ( −1)
2 3
= −15
Function of Two Variables

7. Example
Let f be the function defined by
x
f ( x ,y ) =
x + y −3
Find f ( 2,3 ),f ( 3,2 )
2
f ( 2,3 ) = =1
2+ 3− 3
3 3
f ( 3,2 ) = =
3+ 2− 3 2
Function of Two Variables

8. Example
Find the domain of each function
a. f ( x, y ) = 3 x − 2 y 2
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.
x
b. g( x, y ) =
2x + y − 3
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.
Function of Two Variables

9. Question
Let f be the function defined by
f ( x, y ) =
(
100 1000 + 0.03 x y 2
)
0 .5
( 5 − 0. 2 y ) 2
Find the domain of the function
g(x, y) is defined as long as ( 5 − 0.2y ) 2
≠0
( 5 − 0.2y ) 2 ≠ 0
5 − 0.2 y ≠ 0
y ≠ 25
So the domain is the set of all points (x, y) in
the xy – plane except those on the line y=25
Function of Two Variables

10. Example 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
1 1 1 3
p = 300 − x − y q = 240 − x − y
4 8 8 8
What is the weekly total revenue function R (x,y)?
R ( x, y ) = xp + yq
 1 1   1 3 
= x  300 − x − y ÷+ y  240 − x − y ÷
 4 8   8 8 
1 2 3 2 1
= − x − y − xy + 300 x + 240 y
4 8 4
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Function of Two Variables

11. Recall the Graph of Two Variables
Ex. Plot (4, 2)
Ex. Plot (-2, 1)
Ex. Plot (2, -3)
(4, 2)
(-2, 1)
(2, -3)
Function of Two Variables

12. Graphs of Functions of Two Variables
Three-dimensional coordinate system: (x, y, z)
Ex. Plot (2, 5, 4)
z
4
2 y
5
x
Function of Two Variables

13. Graphs of Functions of Two Variables
Ex. Graph of f (x, y)= 4 – x2 – y2
Function of Two Variables

14. 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
∂f f ( x + h, y ) − f ( x , y )
= lim
∂x h→0 h
∂
provided the limit exits. = f x = f x ( x , y ) = f ( x , y )
∂x
Partial Derivatives

15. 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
∂f f ( x, y + k ) − f ( x , y )
= lim
∂y k →0 k
∂
= f y = f y ( x ,y ) = f ( x ,y )
provided the limit exits. ∂y
Partial Derivatives

16. To get partial derivatives….
 To get fx assume y is a constant and
differentiate with respect to x
Example f ( x, y ) = xy 2 + x 2 y
f x ( x, y ) = (1) y 2 + (2 x) y = y 2 + 2 xy
 To get fy assume x is a constant and
differentiate with respect to y
Example f ( x, y ) = xy 2 + x 2 y
f y ( x, y ) = x(2 y ) + x 2 (1) = 2 xy + x 2
Partial Derivatives

17. Example Compute the first partial derivatives
f ( x, y ) = 3 x 2 y + x ln y
1
f x = 6 xy + ln y f y = 3x + x  ÷
2
 y
Example Compute the first partial derivatives
xy 2 + y
g ( x, y ) = e
2 xy 2 + y
g y = ( 2 xy + 1) e xy 2 + y
gx = y e
Partial Derivatives

18. Example Compute the first partial derivatives
f ( x, y ) = 3 x 2 y − 2 + y 3 .
f x = 6 xy f y = 3x 2 + 3y 2
Example Compute the first partial derivatives
f ( x , y ) = 2x + 3y − 4
fx =2 fy =3
Partial Derivatives

19. The Cobb-Douglas Production Function
1−b
f ( x, y ) = ax y b
• 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
fx is the marginal productivity of labor.
fy is the marginal productivity of capital.
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Partial Derivatives: Application of First Partial Derivatives

20. Example A certain production function is given by
f ( x, y ) = 28 x y units, when x units of
1/ 4 3/ 4
labor and y units of capital are used. Find the
marginal productivity of capital when labor = 81
units and capital = 256 units. 1/ 4
1/ 4 −1/ 4 x
f y = 21x y = 21 ÷
 y
When labor = 81 units and capital = 256 units,
1/ 4
fy  81  3
= 21 ÷ = 21 ÷ = 15.75
 256  4
So 15.75 units per unit increase in capital expenditure.
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Partial Derivatives: Application of First Partial Derivatives

21. Question A certain production function is given by
f ( x, y ) = 28 x y units, when x units of
1/ 4 3/ 4
labor and y units of capital are used. Find the
marginal productivity of labor when labor = 81 units
and capital = 256 units.
 1  −3 / 4 3 / 4
f x = 28  x y = 7 x −3 / 4 y 3 / 4
4
When labor = 81 units and capital = 256 units,
f x = 7(81) −3 / 4 (256) 3 / 4 = 49.78
So 49.78 units per unit increase in labor expenditure.
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Partial Derivatives: Application of First Partial Derivatives

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)
22
Partial Derivatives: Application of First Partial Derivatives

23. Substitute and Complementary Commodities
Two commodities A and B are substitute commodities
if
∂f ∂g
> 0 and >0
∂q ∂p
Two commodities A and B are complementary
commodities if
∂f ∂g
< 0 and <0
∂q ∂p
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Partial Derivatives: Application of First Partial Derivatives

24. Example The demand function for two related commodities
are
x = ae q-p
y = be p-q
The marginal demand functions are
δx = - ae q-p δy = be p-q
δp δp
δx = ae q-p δy = - be p-q
δq δq
Because δx/δq > 0 and δy/δp > 0, the two
commodities are substitute commodities.
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Partial Derivatives: Application of First Partial Derivatives

25. Question
In a survey it was determined that the demand equation for VCRs
is given by
x = f ( p, q ) = 10, 000 − 10 p − e 0.5 q
The demand equation for blank VCR tapes is given by
y = g ( p, q ) = 50, 000 − 4000q − 10 p
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.
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Partial Derivatives: Application of First Partial Derivatives

26. x = f ( p, q ) = 10, 000 − 10 p − e 0.5 q
y = g ( p, q ) = 50, 000 − 4000q − 10 p
∂x ∂y
= −10 = −10
∂p ∂p
∂x ∂y
= −0.5e 0.5 q = −4000
∂q ∂q
Because δx/δq < 0 and δy/δp < 0, the two commodities are
complementary commodities.
Partial Derivatives: Application of First Partial Derivatives

27. Second-Order Partial Derivatives
∂ f
2
∂ ∂2 f ∂
f xx = 2 = ( f x ) f xy = = ( fx )
∂x ∂x ∂y∂x ∂y
∂ f
2
∂ ∂ f 2
∂
f yy = 2 = ( f y ) f yx = = ( fy )
∂y ∂y ∂x∂y ∂x
27
Partial Derivatives: Second-Order Partial Derivatives

28. Example
Find the second-order partial derivatives of the function
f ( x, y ) = 3 x 2 y + x ln y
1
f x = 6 xy + ln y f y = 3x + x  ÷
2
 y
x 1 1
f xx = 6 y f yy =− 2 f xy = 6x + f yx = 6x +
y y y
Example Find the second-order partial derivatives of the function
2
a. f ( x, y ) = 3 x − 2 y
fx =3 f y = −4 y
f xx = 0 f yy = −4 f xy = 0 f yx = 0
Partial Derivatives: Second-Order Partial Derivatives

29. Example Find the second-order partial derivatives of the function
xy 2
f ( x, y ) = e
fx =
∂ xy 2
∂x
e ( )2 xy 2
=ye fy =
∂ xy 2
∂y
e ( )
= 2 xye xy 2
f xx = y e 4 xy 2
f yx = 2 ye xy 2
( 1 + xy )2
f xy = 2 ye xy 2 2
(
+ y 2 xye xy 2
) = 2 ye xy 2
(1 + xy )2
f yy = 2 x e ( xy 2
+ye ( xy 2
))
• 2 xy = 2 xe xy 2
(1 + 2 xy )2
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Partial Derivatives: Second-Order Partial Derivatives

30. Maximum and Minimum of
Functions of Several
Variables
30

31. Relative Extrema of a Function of Two
Variables
Let 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)
for all (x, y) sufficiently close to (a, b).
f (a, b) is a relative minimum of f if f ( x, y ) ≥ f (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.
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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
∂f ∂f
( a, b ) = 0 and ( a, b ) = 0
∂x ∂y
or at least one of the partial derivatives
does not exist.
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Partial Derivatives: Application of Second Partial Derivatives

33. Determining Relative Extrema
1. Find all the critical points by solving the system
f x = 0, f y = 0
2. The 2nd Derivative Test: Compute
D( x, y ) = f xx f yy − f xy
2
D ( a, b) f xx (a, b) Interpretation
+ + Relative min. at (a, b)
+ – Relative max. at (a, b)
– Neither max. nor min. at (a, b)
 saddle point
0 Test is inconclusive
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34. Ex. Determine the relative extrema of the function
f ( x, y ) = 2 x − x 2 − y 2
So the only critical
fx = 2 − 2x = 0 f y = −2 y = 0
point is (1, 0).
f xx = f yy = −2, f xy = 0
D(1, 0) = ( −2 ) ( −2 ) − 0 2 = 4 > 0 and f xx ( 1, 0 ) = −2 < 0
So f (1,0) = 1 is a relative maximum
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35. Application
Ex: The total weekly revenue (in dollars) that Acrosonic realizes
in producing and selling its bookshelf loudspeaker systems is
given by 1 2 3 2 1
R ( x, y ) = − x − y − xy + 300 x + 240 y
4 8 4
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
C ( x, y ) = 180 x + 140 y + 5000
Determine how many assembled units and how many kits
Acrosonic should produce per week to maximize its profit.
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36. P ( x, y ) = R ( x, y ) − C ( x , y )
1 2 3 2 1
= − x − y − xy + 120 x + 100 y − 5000
4 8 4
1 1
Px = − x − y + 120 = 0K ( 1)
2 4
3 1
Py = − y − x + 100 = 0K ( 2 )
4 4
36

37. Substitute in K ( 1)
y = −2 x + 480
Substitute in K ( 2 )
3 1
Py = − ( −2 x + 480 ) − x + 100 = 0
4 4
= 6 x − 1440 − x + 400 = 0
∴ x = 208
Substitute this value into the equation y = −2 x + 480
∴ y = 64
Therefore, P has the critical point (208,64)
37

38. 1 1 3
Pxx = − Pxy = − Pyy = −
2 4 4
2
 1  3   1  5
D ( x, y ) =  − ÷ − ÷−  − ÷ =
 2  4   4  16
Since, D ( 208, 64 ) > 0 and Pxx ( 208, 64 ) < 0 , the
point (208,64) is a relative maximum of P.
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39. Lagrange Multipliers
Reading: Mizrahi and Sullivan, 8th ed., 2004, Wiley
Chapter:17.5
39

40. Method of Lagrange Multipliers
A method to find the local minimum and maximum of a
function with two variables subject to conditions or
constraints on the variables involved.
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
.
( x0 , y0 )
Form the function
F ( x, y , λ ) = f ( x, y ) + λ g ( x, y )
40

41. Then there is a value of λ such that ( x0 , y0 , λ ) is a solution
of the system of equations
∂F ∂f ∂g
= +λ = 0 L ( 1)
∂x ∂x ∂x
∂F ∂f ∂g
= +λ = 0 L( 2)
∂y ∂y ∂y
∂F
= g ( x, y ) = 0 L ( 3)
∂λ
provided all the partial derivatives exists.
41

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
z = f ( x, y )
subject to the constraint g ( x, y ) = 0
Step 2: Construct the function F:
F ( x, y , λ ) = f ( x, y ) + λ g ( x, y )
42

43. Step 3: Set up the system of equations
∂F
= 0 L ( 1)
∂x
∂F
= 0 L( 2)
∂y
∂F
= g ( x, y ) = 0 L ( 3 )
∂λ
Step 4: Solve the system of equations for x, y and λ .
Step 5: Test the solution ( x0 , y0 , λ ) to determine
maximum or minimum point.
43

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 z = f ( x, y ) at each solution ( x0 , y0 , λ )
found in Step 5.
44

45. Example:
Find the minimum of
f(x,y) = 5x2 + 6y2 - xy
subject to the constraint
x+2y = 24
Solution:
F(x,y, λ) = 5x2 + 6y2 - xy + λ(x + 2y - 24)
Fx = δF = 10x - y + λ ; Fxx = 10
δx
Fy = δF = 12y - x + 2λ ; Fyy = 12
δy
Fλ = δF = x + 2y - 24 ; Fxy = -1
δλ
45

46. The critical point,
10x - y + λ = 0
12y - x + 2λ= 0
x + 2y - 24= 0
The solution of the system is x = 6, y = 9, λ = -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
46

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
P ( x, y ) = x 2 + 3 xy − 6 y
to maximize profit, how many engines of each type should be
produced if there must be a total of 42 engines produced?
47

48. Maximize z = P ( x, y ) = x + 3xy − 6 y
2
Subject to constraint g ( x, y ) = x + y − 42 = 0
F ( x, y , λ ) = P ( x , y ) + λ g ( x , y )
= x 2 + 3xy − 6 y + λ ( x + y − 42 )
∂F
= 2 x + 3 y + λ = 0 L ( 1) ; Fxx = 2
∂x
∂F
= 3x − 6 + λ = 0 L ( 2 ) ; Fyy = 0
∂y
∂F
= x + y − 42 = 0 L ( 3) ; Fxy = 3
∂λ
The solution of the system is x = 33 y = 9 λ = −93.
48

49. Fxx = 2 > 0
D* = (2)(0) − (3) 2 = −9 < 0
The test in inconclusive.
49