Introductory maths analysis chapter 17 official

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Introductory maths analysis chapter 17 official

  1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 17Chapter 17 Multivariable CalculusMultivariable Calculus
  2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
  3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
  4. 4. ©2007 Pearson Education Asia • To discuss functions of several variables and to compute function values. • To compute partial derivatives. • To develop the notions of partial marginal cost, marginal productivity, and competitive and complementary products. • To find partial derivatives of a function defined implicitly. • To compute higher-order partial derivatives. • To find the partial derivatives of a function by using the chain rule. Chapter 17: Multivariable Calculus Chapter ObjectivesChapter Objectives
  5. 5. ©2007 Pearson Education Asia • To apply the second-derivative test for a function of two variables. • To find critical points for a function. • To develop the method of least squares. • To compute double and triple integrals. Chapter 17: Multivariable Calculus Chapter ObjectivesChapter Objectives
  6. 6. ©2007 Pearson Education Asia Functions of Several Variables Partial Derivatives Applications of Partial Derivatives Implicit Partial Differentiation Higher-Order Partial Derivatives Chain Rule Maxima and Minima for Functions of Two Variables 17.1) 17.2) 17.3) Chapter 17: Multivariable Calculus Chapter OutlineChapter Outline 17.4) 17.5) 17.6) 17.7)
  7. 7. ©2007 Pearson Education Asia Lagrange Multipliers Lines of Regression Multiple Integrals 17.8) 17.9) 17.10) Chapter 17: Multivariable Calculus Chapter OutlineChapter Outline
  8. 8. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.1 Functions of Several Variables17.1 Functions of Several Variables Example 1 – Functions of Two Variables • A function can involve 2 or more variables, e.g. ( ) 22 2 , yx yxfz − == a. is a function of two variables. Because the denominator is zero when y = 2, the domain of f is all (x, y) such that y ≠ 2. b. h(x, y) = 4x defines h as a function of x and y. The domain is all ordered pairs of real numbers. c. z2 = x2 + y2 does not define z as a function of x and y. ( ) 2 3 , − + = y x yxf
  9. 9. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.1 Functions of Several Variables Example 3 – Graphing a Plane Sketch the plane Solution: The plane intersects the x-axis when y = 0 and z = 0. .632 =++ zyx
  10. 10. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.1 Functions of Several Variables Example 5 – Sketching a Surface Sketch the surface Solution: z = x2 is a parabola. .2 xz =
  11. 11. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.2 Partial Derivatives17.2 Partial Derivatives • Partial derivative of f with respect to x is given by • Partial derivative of f with respect to y is given by ( ) ( ) ( ) h yxfyhxf yxf h x ,, lim, 0 −+ = → ( ) ( ) ( ) h yxfhyxf yxf h y ,, lim, 0 −+ = →
  12. 12. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.2 Partial Derivatives Example 1 – Finding Partial Derivatives If f(x, y) = xy2 + x2 y, find fx(x, y) and fy(x, y). Also, find fx(3, 4) and fy(3, 4). Solution: The partial derivatives are Thus, the solutions are ( ) ( ) ( ) xyyyxyyxfx 221, 22 +=+= ( ) ( ) ( ) 22 212, xxyxyxyxfy +=+= ( ) ( ) 334,3 404,3 = = xy x f f
  13. 13. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.2 Partial Derivatives Example 3 – Partial Derivatives of a Function of Three Variables If find Solution: By partial differentiating, we get ( ) ,,, 322 zzyxzyxf ++= ( ) ( ) ( ).,,and,,,,, zyxfzyxfzyxf zyx ( ) xzyxfx 2,, = ( ) yzzyxfy 2,, = ( ) 22 3,, zyzyxfz +=
  14. 14. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.2 Partial Derivatives Example 4 – Finding Partial Derivatives If find Solution: By partial differentiating, we get ( ) ,,,, 22 tsrt rsu utsrgp + == ( ) .and, 1,1,1,0t p t p s p ∂ ∂ ∂ ∂ ∂ ∂ ( )( ) ( )( ) ( ) ( ) ( )22 2 222 22 2 srtt srtru tsrt strsurutsrt s p + − = + −+ = ∂ ∂ ( ) ( ) ( ) ( )222 2 2222 2 2 tsrt srtrsu s p srttsrtrsu t p + + −= ∂ ∂ ⇒++−= ∂ ∂ − ( ) 0 1,1,1,0 = ∂ ∂ t p
  15. 15. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.3 Applications of Partial Derivatives17.3 Applications of Partial Derivatives Example 1 – Marginal Costs • Interpretation of rate of change: A company manufactures two types of skis, the Lightning and the Alpine models. Suppose the joint-cost function for producing x pairs of the Lightning model and y pairs of the Alpine model per week is where c is expressed in dollars. Determine the marginal costs ∂c/∂x and ∂c/∂y when x = 100 and y = 50, and interpret the results. fixed.heldiswhentorespectwithofchangeofratetheis fixed.heldiswhentorespectwithofchangeofratetheis xyz y z yxz x z ∂ ∂ ∂ ∂ ( ) 6000857507.0, 2 +++== yxxyxfc
  16. 16. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.3 Applications of Partial Derivatives Example 1 – Marginal Costs Solution: The marginal costs are Thus, and ( ) ( ) ( ) 85$ 89$7510014.0 50,100 50,100 = ∂ ∂ =+= ∂ ∂ y c x c 85and7514.0 = ∂ ∂ += ∂ ∂ y c x x c
  17. 17. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.3 Applications of Partial Derivatives Example 3 – Marginal Productivity A manufacturer of a popular toy has determined that the production function is P = √(lk), where l is the number of labor-hours per week and k is the capital (expressed in hundreds of dollars per week) required for a weekly production of P gross of the toy. (One gross is 144 units.) Determine the marginal productivity functions, and evaluate them when l = 400 and k = 16. Interpret the results.
  18. 18. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.3 Applications of Partial Derivatives Example 3 – Marginal Productivity Solution: Since , Thus, ( ) lk l k P lk k klk l P 2 and 22 1 2/1 = ∂ ∂ == ∂ ∂ − ( ) 2/1 lkP = 2 5 and 10 1 16,40016,400 = ∂ ∂ = ∂ ∂ ==== klkl k P l P
  19. 19. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.4 Implicit Partial Differentiation17.4 Implicit Partial Differentiation Example 1 – Implicit Partial Differentiation • We will look into how to find partial derivatives of a function defined implicitly. If , evaluate when x = −1, y = 2, and z = 2. Solution: Using partial differentiation, we get 02 2 =+ + y yx xz x z ∂ ∂ ( ) ( ) ( ) ( ) ( ) 0 2 02 0 22 2 2 ≠ + −= ∂ ∂ =−++ ∂ ∂ + ∂ ∂ = ∂ ∂ +      +∂ ∂ z yxx yz x z xzyxz x z yxxz x y xyx xz x ( ) 2 2,2,1 = ∂ ∂ −x z
  20. 20. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.5 Higher-Order Partial Derivatives17.5 Higher-Order Partial Derivatives Example 1 – Second-Order Partial Derivatives • We obtain second-order partial derivatives of f as Find the four second-order partial derivatives of Solution: yyyyxyyx yxxyxxxx ffff ffff )(meansand)(means )(meansand)(means ( ) ., 222 yxyxyxf += ( ) ( ) ( ) xyxyxfyyyxf xyxyyxf xyxx x 42,and22, 22, 2 2 +=+= += ( ) ( ) ( ) xyxyxfxyxf yxxyxf yxyy y 42,and2, 2, 2 22 +== +=
  21. 21. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.5 Higher-Order Partial Derivatives Example 3 – Second-Order Partial Derivative of an Implicit Function Determine if Solution: By implicit differentiation, Differentiating both sides with respect to x, we obtain Substituting , 2 2 x z ∂ ∂ .2 xyz = ( ) ( ) 0 2 2 ≠= ∂ ∂ ⇒ ∂ ∂ = ∂ ∂ z z y x z xy x z x x z yz x z yz xx z x ∂ ∂ −= ∂ ∂ ⇒      ∂ ∂ =      ∂ ∂ ∂ ∂ −− 2 2 2 1 2 1 2 1 z y x z 2 = ∂ ∂ 0 422 1 3 2 2 2 2 ≠−=      −= ∂ ∂ − z z y z y yz x z
  22. 22. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.6 Chain Rule17.6 Chain Rule • If f, x, and y have continuous partial derivatives, then z is a function of r and s, and s y y z s x x z s z r y y z r x x z r z ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ and
  23. 23. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.6 Chain Rule Example 1 – Rate of Change of Cost For a manufacturer of cameras and film, the total cost c of producing qC cameras and qF units of film is given by The demand functions for the cameras and film are given by where pC is the price per camera and pF is the price per unit of film. Find the rate of change of total cost with respect to the price of the camera when pC = 50 and pF = 2. 900015.030 +++= FFC qqqqcc FCF F ppq ppc qc 4002000and 9000 −−==
  24. 24. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.6 Chain Rule Example 1 – Chain Rule Example 3 – Chain Rule Solution: By the chain rule, Thus, ( ) ( )( )11015.0 9000 015.030 2 −++        − += ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ C FC F C F FC C CC q pp q p q p c p q q c p c 2.123 2,50 −≈ ∂ ∂ == FC ppCp c a. Determine ∂y/∂r if Solution: By the chain rule, ( ) ( ) .3and6ln 642 srxxxy +=+= ( ) ( )      ++ + += ∂ ∂ ∂ ∂ = ∂ ∂ 6ln 6 2 312 4 4 4 5 x x x srx r x x y r y
  25. 25. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.6 Chain Rule Example 3 – Chain Rule b. Given that z = exy , x = r − 4s, and y = r − s, find ∂z/∂r in terms of r and s. Solution: By the chain rule, Since x = r − 4s and y = r − s, ( ) xy eyx r y y z r x x z r z += ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ = ∂ ∂ ( ) 22 45 4 52 srsr sry xrx esr r z +− −= −= −= ∂ ∂
  26. 26. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions17.7 Maxima and Minima for Functions of Two Variablesof Two Variables • Relative maximum at the point (a, b) is shown as RULE 1 Find relative maximum or minimum when RULE 2 Second-Derivative Test for Functions of Two Variables Let D be the function defined by 1. If D(a, b) > 0 and fxx(a, b) < 0, relative maximum at (a, b); 2. If D(a, b) > 0 and fxx(a, b) > 0, relative minimum at (a, b); 3. If D(a, b) < 0, then f has a saddle point at (a, b); 4. If D(a, b) = 0, no conclusion. ( ) ( )yxfbaf ,, ≥ ( ) ( )   = = 0, 0, yxf yxf y x ( ) ( ) ( ) ( )( ) .,,,, 2 yxfyxfyxfyxD xyyyxx −=
  27. 27. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 1 – Finding Critical Points Find the critical points of the following functions. a. Solution: Since we solve the system and get b. Solution: Since we solve the system and get and ( ) 13522, 22 +−+−+= yxxyyxyxf ( ) ( ) ,0322,and0524, =−+−==+−= yxyxfyxyxf yx    = −= 2 1 1 y x ( ) lkklklf −+= 33 , ( ) ( ) ,03,and03, 22 =−==−= lkklfklklf kl    = = 0 0 k l . 3 1 3 1    = = k l
  28. 28. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 1 – Finding Critical Points c. Since we solve the system and get ( ) ( )1001002,, 22 −+−+++= yxzyxyxzyxf ( ) ( ) ( ) 0100,, 02,, 04,, =+−−= =−+= =−+= yxzyxf zyxzyxf zyxzyxf z y x      = = = 175 75 25 z y x
  29. 29. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 3 – Applying the Second-Derivative Test Examine f(x,y) = x3 + y3 − xy for relative maxima or minima by using the second derivative test. Solution: We find critical points, which gives (0, 0) and (1/3, 1/3). Now, Thus, D(0, 0) < 0  no relative extremum at (0, 0). D(1/3,1/3)>0 and fxx(1/3,1/3)>0 relative minimum at (1/3,1/3) Value of the function is ( ) ( ) 03,and03, 22 =−==−= xyyxfyxyxf yx ( ) ( ) ( ) 1,6,6, −=== yxfyyxfxyxf xyyyxx ( ) ( )( ) ( ) 136166, 2 −=−−= xyyxyxD ( ) ( ) ( ) ( )( ) 27 1 3 1 3 13 3 13 3 1 3 1 3 1 , −=−+=f
  30. 30. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 5 – Finding Relative Extrema Examine f(x, y) = x4 + (x − y)4 for relative extrema. Solution: We find critical points at (0,0) through D(0, 0) = 0  no information. f has a relative (and absolute) minimum at (0, 0). ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0,12,02112, 04,044, 222 333 =−==−+= =−−==−+= yxfyxyxfyxxyxf yxyxfyxxyxf xyyyxx yx
  31. 31. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 7 – Profit Maximization A candy company produces two types of candy, A and B, for which the average costs of production are constant at $2 and $3 per pound, respectively. The quantities qA, qB (in pounds) of A and B that can be sold each week are given by the joint-demand functions where pA and pB are the selling prices (in dollars per pound) of A and B, respectively. Determine the selling prices that will maximize the company’s profit P. ( ) ( )BAB ABA ppq ppq 29400 400 −+= −=
  32. 32. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.7 Maxima and Minima for Functions of Two Variables Example 7 – Profit Maximization Solution: The total profit is given by The profits per pound are pA − 2 and pB − 3, The solution is pA = 5.5 and pB = 6. Since ∂2 P/∂p2 A< 0, we indeed have a maximum. ( ) ( ) 01342and0122 32 =+−= ∂ ∂ =−+−= ∂ ∂ −+−= BA B BA A BBAA pp p P pp p P qpqpP 8001600800 2 2 2 2 2 = ∂∂ ∂ −= ∂ ∂ −= ∂ ∂ ABBA pp P p P p P ( ) 06,5.5 >D
  33. 33. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.8 Lagrange Multipliers17.8 Lagrange Multipliers Example 1 – Method of Lagrange Multipliers • Lagrange multipliers allow us to obtain critical points. • The number λ0 is called a Lagrange multiplier. Find the critical points for z = f(x,y) = 3x − y + 6, subject to the constraint x2 + y2 = 4. Solution: Constraint Construct the function Setting , we solve the equations to be ( ) 04, 22 =−+= yxyxg ( ) ( ) ( ) ( )463,,,, 22 −+−+−=−= yxλyxyxgλyxfλyxF 0=== λyx FFF 4 10 and 2 1 , 2 3 ±=−== λ λ y λ x      =+−− =−− =− 04 021 023 22 yx λy λx
  34. 34. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.8 Lagrange Multipliers Example 3 – Minimizing Costs Suppose a firm has an order for 200 units of its product and wishes to distribute its manufacture between two of its plants, plant 1 and plant 2. Let q1 and q2 denote the outputs of plants 1 and 2, respectively, and suppose the total-cost function is given by How should the output be distributed in order to minimize costs? ( ) .2002,, 2 221 2 121 +++== qqqqλqqfc
  35. 35. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.8 Lagrange Multipliers Example 3 – Minimizing Costs Solution: We minimize c = f(q1, q2), given the constraint q1 + q2 = 200. ( ) ( )2002002,, 21 2 221 2 121 −+−+++= qqλqqqqλqqF          =+−−= ∂ ∂ =−+= ∂ ∂ =−+= ∂ ∂ 0200 02 04 21 21 2 21 1 qq λ F λqq q F λqq q F 150,50 21 == qq
  36. 36. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.8 Lagrange Multipliers Example 5 – Least-Cost Input Combination Find critical points for f(x, y, z) = xy+ yz, subject to the constraints x2 + y2 = 8 and yz = 8. Solution: Set We obtain 4 critical points: (2, 2, 4) (2,−2,−4) (−2, 2, 4) (−2,−2,−4) ( ) ( ) ( )88,,,, 2 22 121 −−−+−+= yzλyxλyzxyλλzyxF          =+−= =+−−= =−= =−−+= =−= 08 08 0 02 02 2 1 22 2 21 1 yzF yxF λyyF λzλyzxF λxyF λ λ z y x           = =+ = =−−+ = y z yx λ λzλyzx λ x y 8 8 1 02 2 22 2 21 1
  37. 37. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.9 Lines of Regression17.9 Lines of Regression 2 11 2 1111 2       −             −            = ∑∑ ∑∑∑∑ == ==== n i i n i i n i ii n i i n i i n i i xxn yxxyx a 2 11 2 111       −             − = ∑∑ ∑∑∑ == === n i i n i i n i i n i ii n i i xxn yxyxn b
  38. 38. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.9 Lines of Regression Example 1 – Determining a Linear-Regression Line By means of the linear-regression line, use the following table to represent the trend for the index of total U.S. government revenue from 1995 to 2000 (1995 = 100).
  39. 39. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.9 Lines of Regression Example 1 – Determining a Linear-Regression Line Solution: Perform the arithmetic ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 83.9 21916 7362127486 3.88 21916 27482173691 2 2 ≈ − − = ≈ − − = b a xy 83.93.88ˆ +=
  40. 40. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.10 Multiple Integrals17.10 Multiple Integrals Example 1 – Evaluating a Double Integral • Definite integrals of functions of two variables are called (definite) double integrals, which involve integration over a region in the plane. Find Solution: ( ) .12 1 1 1 0 dxdyx x ∫ ∫− − + ( ) [ ] 3 2 23 2 2 12 1 1 231 1 1 0 1 1 1 0 =      ++−=+= + −− − − − ∫ ∫ ∫ x xx dxyxy dxdyx x x
  41. 41. ©2007 Pearson Education Asia Chapter 17: Multivariable Calculus 17.10 Lines of Regression Example 3 – Evaluating a Triple Integral Find Solution: . 1 0 0 0 dxdydzx x yx ∫∫ ∫ − [ ] 8 1 82 1 0 41 0 0 2 2 1 0 0 0 1 0 0 0 =      =      −= = ∫ ∫∫∫∫ ∫ − − x dx xy yx dxdyxzdxdydzx x x yx x yx

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