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- 1. Partial Differentiation &Application Week 9
- 2. Contents:1. Function with two variables2. First Partial Derivatives3. Applications of First Partial Derivatives Cob-Douglas Production Function Substitute and Complementary Commodities1. Second Partial Derivatives2. 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 VariablesA 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 10 Function of Two Variables
- 11. Recall the Graph of Two VariablesEx. 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 firstpartial derivative of f with respect to y at apoint (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 xExample 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 yExample 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 yExample 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 2Example 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. 19 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. 20 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. 21 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 CommoditiesTwo commodities A and B are substitute commoditiesif ∂f ∂g > 0 and >0 ∂q ∂pTwo commodities A and B are complementarycommodities if ∂f ∂g < 0 and <0 ∂q ∂p 23 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. 24 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. 25 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 ∂qBecause δx/δq < 0 and δy/δp < 0, the two commodities arecomplementary 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 1f xx = 6 y f yy =− 2 f xy = 6x + f yx = 6x + y y yExample 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 29 Partial Derivatives: Second-Order Partial Derivatives
- 30. Maximum and Minimum of Functions of Several Variables 30
- 31. Relative Extrema of a Function of TwoVariablesLet 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 off then the points are absolute extrema. 31 Partial Derivatives: Application of Second Partial Derivatives
- 32. Critical Point of fA critical point of f is a point (a, b) in thedomain of f such that both ∂f ∂f ( a, b ) = 0 and ( a, b ) = 0 ∂x ∂yor at least one of the partial derivativesdoes not exist. 32 Partial Derivatives: Application of Second Partial Derivatives
- 33. Determining Relative Extrema1. Find all the critical points by solving the system f x = 0, f y = 02. The 2nd Derivative Test: Compute D( x, y ) = f xx f yy − f xy 2D ( 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 33
- 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 = 0D(1, 0) = ( −2 ) ( −2 ) − 0 2 = 4 > 0 and f xx ( 1, 0 ) = −2 < 0 So f (1,0) = 1 is a relative maximum 34
- 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. 35
- 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 1Px = − x − y + 120 = 0K ( 1) 2 4 3 1Py = − 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 = 208Substitute this value into the equation y = −2 x + 480 ∴ y = 64Therefore, 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 16Since, D ( 208, 64 ) > 0 and Pxx ( 208, 64 ) < 0 , thepoint (208,64) is a relative maximum of P. 38
- 39. Lagrange MultipliersReading: Mizrahi and Sullivan, 8th ed., 2004, WileyChapter:17.5 39
- 40. Method of Lagrange MultipliersA method to find the local minimum and maximum of afunction with two variables subject to conditions orconstraints on the variables involved.Suppose that, subject to the constraint g(x,y)=0, the functionz=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 solutionof 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 MultipliersStep 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 ) = 0Step 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 ) ∂λStep 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)2If D* > 0 ⇒ Fxx < 0 ∴ maximum point Fxx > 0 ∴ minimum point D* ≤ 0 ⇒ Test is inconclusiveStep 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 - xysubject to the constraint x+2y = 24Solution: 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= 0The solution of the system is x = 6, y = 9, λ = -51 D*=(10)(12)-(-1)2=119>0 Fxx = 10>0We 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 ∂λThe solution of the system is x = 33 y = 9 λ = −93. 48
- 49. Fxx = 2 > 0 D* = (2)(0) − (3) 2 = −9 < 0The test in inconclusive. 49

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