Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : WHAT IS THE DIFFERENTIATION.

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1 Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : WHAT IS THE DIFFERENTIATION. John Wiley and Sons 13

2 Slopes/rates of change Recall linear functions For linear functions slope is the change in y per unit increase in x Slope is the rate of change The slope or rate of change is the same over any interval of any size See the following example

3 Example: calculate the rate of change for a linear function Suppose the distance, (x m) that a ball after y (seconds) is given by the equation y = x y (distance in m) Slope is the rate of change Δx = 4-1=3 y 3 = = = 1 x 3 The average speed of the ball is 1 meter per second, between the 1 st and 4 th sec Δy = 4-1 = 3 x (time in seconds)

4 Example: the rate of change for a linear function.. continued Now take a different interval, 1 to 3 sec y (distance in m) Average speed 5 is 1 meter per second Δx =3-1= Slope is the rate of change = y x Δy = 3-1= = = The slope of this linear function (speed) is 1 m/s when measured over any interval of any length! 1 x (time in seconds)

5 The slope of a curve varies along the curve Suppose the distance, (x m) that a ball after y (seconds) is given by the equation y = x y (distance in m) x (time in seconds) The slope varies along the curve.

6 How do find the rate of change (speed) at x = 1? Start by getting the average slope over progressively smaller intervals where one end is fixed at x = 1. y (distance in m) x (time in seconds) The average slope over the smallest possible interval at x = 1approximates the slope at x = 1.

7 What is the rate of change (speed) at x = 1 Start by getting the average slope over progressively smaller intervals, where one end is fixed at x = y (distance in m) x (time in seconds) The average slope over the smallest possible interval at x = 1 approximates the slope at x = 1.

8 Average slope of the curve between x = 1 and x = 3 1 Slope of chord BC 9 y 8 m = = 8 = 4 x 7 Slope or the average speed between 1st and 3rd second is 4 m/sec C y (distance in m) B Δx = 3-1= Δy = 9-1= x (time in seconds)

9 Average slope of the curve between x = 1 and x =.5 1 Slope of chord BC m = 8 = Slope or the average speed between 1 and.5 seconds is 3.5 m/sec C y (distance in m) Δy = 6.5-1=5.5 B Δx =.5-1= x (time in seconds)

10 Average slope of the curve between x = 1 and x = 1 Slope of chord BC m = = Slope or the average speed between the 1st and nd seconds is 3 m/sec C y (distance in m) B Δx = - 1=1 Δy = 4-1= x (time in seconds)

11 Average slope of the curve between x = 1 and x = Slope of chord BC m = 8 = Slope or the average speed between 1 and 1.5 seconds is.5 m/sec B Δy =.5-1=1.5 Δx = 1.5-1= C

12 Average slope of the curve between x = 1 and x = Slope of chord BC m = 8 = Slope or the average speed between 1 and 1.1 seconds is.1 m/sec B C Δy = 1.1-1= Δx = 1.1-1=.1

13 The slope of a curve at x = 1 1 Slope of chord at B m = = Zero divided zero is not defined! So. we need another method to find the slope at a point B

14 Summary, so far y = x is the distance travelled, y after x seconds Time interval: between Average speed 1 st and 3 rd second 4 m/s 1 and.5 seconds 3.5 m/s 1 st and nd second 3 m/s Slope (average speed) is approaching the value as the interval Δx is becoming progressively smaller towards x = 1 1 and 1. 5 second.5 m/s 1 and 1.1 second.1 m/s at 1 second Not defined???

15 Find slope of a curve at a point, graphically Slope of the 1curve at B = slope of the tangent at B Measure the 9slope (m) of the tangent line m = h d m = = B d =. But the accuracy of graphical methods is limited! h =

16 The slope of a curve at a point is determined analytically by differentiation. Method: given the equation of the curve, y = x First, find the derivative of y = x d y = dx (see Worked Example 6.1 for proof ) Then evaluate the derivative at the given point At x = 1, x dy dx = x = (1) = at x = 1

17 The slope of a curve at a point, graphically and analytically Slope of the curve 1 at B = slope of the tangent at B Graphically, measure 9 the slope (m) of the tangent line m = h d m = 4 = Analytically, curve y = x At x = 1, the slope, dy dx m = x 1 = (1) = = at x = B d =. h =

18 Summary: Slope of chord = 4, 1 Slope the average of chord slope =3.5, of 9 the average curve between slope of 8 Slope the of x = chord curve 1 and =3, between x = 3 7 Slope the of average chord x = 1 =.5, and slope x = of.5 6 the curve between Slope By differentiation the of curve average 5 slope of the x = 1 between and x = x slope = 1 is of the curve same 4 at as the x = slope 1 x is = d1 of yand the x = 1.5 tangent at dx =1 = 3 x = at x =

19 The slope of a curve at a point is determined analytically by differentiation Method consists of two simple steps. First Given the curve, y = f(x), find its derivative, Next dy dx Evaluate the derivative at the given point The slope of a curve at a point is also described as the instantaneous rate of change

20 Worked Example 6.1 Figure 6.3: Equation of the curve y = x At x = -1, dy/dx = (-1) = - At x = 1.5, y =.5, dy/dx = (1.5) =3 Equation for slope of curve, dy/dx = x y (ii) y = -x x (i) y = 3x Figure 6.3: Slope of curve y = x at x = -1 and x = 1.5

21 Worked Example 6.1 Figure 6.3: Equation of the curve y = x Equation for slope of curve, dy/dx = x At x = -1, dy/dx = (-1) = - At x = 1.5, y =.5, dy/dx = (1.5) Figure 6.3: Slope of curve y = x at x = -1 and x = 1.5

22 Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : MAX MIN METHOD. John Wiley and Sons 13

23 PowerPoint slides introduce the following topics The Power Rule for differentiation Demonstrate that (i) the slope of the curve is positive when y is increasing (ii) the slope of the curve is positive when y is increasing Use first derivative to locate stationary points Worked Example 6.17, Figure 6.16 Use second derivative to determine the nature of stationary point. Worked Example 6.19: Derived Curves

24 The Power Rule for Differentiation If y = x n then dy dx = nx n 1 You could describe the power rule very informally: giving y dy dx = x n = nx n 1 n drops down: then the power (index) looses 1

25 Power Rule for Differentiation: Examples Function Derivative of function y = x n dy dx = nx n 1 y y = x 3 = x. 8 dy dx dy dx = 3x =. 8x =. 8x. 8 1.

26 When is the derivative positive or negative? When y is increasing, the slope of the curve at any point is positive When y is decreasing the slope of the curve at any point is negative y All tangents have positive slopes y = f(x) y y decreasing All tangents have negative slopes y increasing x x Hence slope, d y > dx dy/dx is positive when y is increasing Hence, slope, d y < dx dy/dx is negative when y is decreasing

27 OPTIMIZATION Finding maximum and minimum points

28 Slope is zero at maximum and at minimum points Tangents drawn at turning points are horizontal: slope = maximum points At turning points slope = dy/dx = d b a c At turning points slope =, dy/dx = minimum point Figure 6.1

29 Locating Turning Points Use the fact that slope is zero at turning points to locate the turning point(s), hence Step 1. Derive the equation for slope by differentiation Step. Substitute slope = in the equation for slope and solve for x. The solution is the x co-ordinate of the turning point(s) Step b. Find the y coordinate by substituting the value of the x coordinate into the equation of the curve

30 Example: locate the turning point of a curve Worked Example 6.16(a)(i) Given the function: y = 3x - 18x + 34 (a) Sketch the function over the interval x = to 5, hence estimate the turning point graphically (b) Use differentiation to locate the turning point

31 (a)(i) Find the turning point graphically y = 3x - 18x + 34 The sketch is given below: y = 3x - 18x + 34 y y = x = 3 Similar to Figure 6.13 Minimum at x = 3, y = 7 x

32 (a)(i) Use differentiation to find the turning point for y = 3x - 18x + 34 Determine the equation for slope by differentiation Step 1. Equation for slope dy dx Equation for slope is dy/dx = 6x 18 = 3(x) 18(1) + Step. At turning points, dy/dx = = 6x 18 x = 3 The x co-ordinate of the turning point is x = 3. Step b.when x = 3, y = 3(3) - 18(3) + 34 = 7. The y co-ordinate of the turning point is y = 7 Turning point at x = 3, y = 7

33 Worked Example 6.16(a)(ii): Locate the turning point an AC curve Given the average cost function: AC = 3Q - 18Q + 34 (a) Sketch the function over the interval Q = to 5, hence estimate the turning point (b) Use differentiation to locate the turning point

34 (a)(ii) AC = 3Q - 18Q + 34 Note: this curve has the same equation as the previous example The sketch is given below: AC 18 AC = 3Q - 18Q AC = Q = 3 Figure 6.13 Minimum AC at AC = 3, AC = 7 Q

35 (a)(ii) Use differentiation to find the turning point for the AC curve: AC = 3Q 18Q + 34 Determine the equation for slope by differentiation Step 1. d( AC) dq = 3(Q) 18(1) + = 6Q 18 Step. At turning points, d(ac)/dq = = 6Q 18 Q = 3 The Q co-ordinate of the turning point is Q = 3. Step b.when Q = 3, AC = 3(3) - 18(3) + 34 = 7. The Q co-ordinate of the turning point is AC = 7 Turning point at Q = 3, AC = 7

36 NATURE OF STATIONARY POINTS Use second derivatives to determine the nature of stationary points

37 Why is the second derivative positive when evaluated at a minimum point? Take any point before, at and after the minimum point For example, points A, B and C in Figure 6.15(b) A small tangent line is drawn to show the slope at points A, B and C. The value of the slope at A, B and C is -.1, and.4 respectively These points demonstrate that the value of slope, dy/dx, increases as we move through a minimum point increasing values Hence, the rate of change of slope is positive (y ʹ < ) (yʹ > ).4 A -.1 (yʹ = ) B Slopes -.1, and.4 at A, B and at C respectively Figure 6.15(b) C d dx dy dx = d y dx > yʹʹ is positive at a minimum point Minimum

38 Why is the second derivative negative when evaluated at a maximum point? Take any point before, at and after the maximum point For example, points A, B and C in Figure 6.15(a) A small tangent line is drawn to show the slope at points A, B and C. The value of the slope at A, B and C is., and -.1 respectively These points demonstrate that the value of slope, dy/dx, decreases as we move through a maximum point. (yʹ > ) A 1 Slopes., and -.1 at A, B and at C respectively Figure 6.15(a) B (yʹ = ) C (y ʹ < ). -.1 decreasing values Hence, the rate of change of slope is negative d dx dy dx = d y dx < yʹʹ is negative at a maximum point Maximum

39 Summary: Use First and Second Derivatives to find Maximum and Minimum Points Tangents drawn at turning points are horizontal: slope = maximum points Maximum points slope, dy/dx = and d y/dx is negative d b a c Minimum points slope, dy/dx = and d y/dx is positive minimum point Figure 6.1

40 Summary of Method for finding and determining the maximum and minimum Points Step 1. Determine the first and second derivatives: yʹ and yʹʹ Step a. Solve d y = dx to find the x co-ordinate of the turning point(s) Step b. Calculate the y coordinate(s) of the turning point(s) Step 3. Slope Test. Evaluate slope at points immediately before and after turning point: About a minimum: slope changes from negative to zero to positive About a maximum: slope changes from positive to zero to negative Step 3. Second derivative Test. At the turning point, evaluate the second derivative, yʹʹ The point is a minimum if yʹʹ is positive The point is a maximum if yʹʹ is negative

41 Use differentiation to find and determine the nature of the turning point for the curve: y = 3x - 18x + 34 Step 1. Determine the first and second derivatives: yʹ and yʹʹ dy dx = 6x 18; d y dx = 6 When the second derivative is a constant there is only one turning point Step a. Solve the equation dy/dx = to locate turning points = 6x 18. Hence x = 3 This is the x co-ordinate of the turning point Step b. When x = 3, y = 3(3) - 18(3) + 34 = 7. Step 3. The second derivative is 6: The y co-ordinate of the turning point is y = 7 yʹʹ is a positive constant, so the turning point is a minimum The minimum point is x = 3, y = 7

42 Worked Example 6.19: Turning points for The method is set out as follows 3 y = 3x 1. x Step 1. Get the first and second derivatives: yʹ and yʹʹ y = 6x. 3x y = 6. 6x Step a. Solve yʹ =, to locate turning points = 6x.3x = x (6.3x) x = (6 -.3x) = x = Turning points at x =, y = and x =, y = 4 Step b. when x =, when x =, y y = 3() = 3().1() 3 =.1() 3 = 4 Continued

43 3 Worked Example 6.19: y = 3x 1. x Continued Step 1. The first and second derivatives: y = 6x. 3x y = 6. 6x Step Turning points were found at (i) at x =, y = and (ii) at x =, y = 4 Step 3: Use slope test or second derivative test to determine the nature of each turning point: Here, we use the second derivative test y = 6. 6x At x =, yʹʹ = 6 -.6() = 6: So, yʹʹ is positive Therefore a minimum at x =, y = At x =, yʹʹ = 6 -.6() = -6 yʹʹ is negative: hence x =, y = 4 is a maximum point.

44 Figure 6.18(a) y = 3x 1. x y maximum= 4 minimum = x

45 y yʹ y = 3x 1. x Minimum at (,) 3 Maximum at (, 4) y = 6x. 3x yʹ is zero when y is a minimum yʹ is zero when y is a maximum

46 45 4 maximum= 4 y minimum = dy dx 4 yʹ = yʹ = d y dx Figure Derived curves yʹʹ = 6 yʹʹ = 6

47 Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : MAX AND MIN APPLICATIONS. John Wiley and Sons 13

48 Worked Example 6.1, Figure 6.3. Determine the maximum Revenue by differentiation Worked Example 6., Figure 6.4: Determine maximum profit MR = MC when profit is maximized Figure 6.9 Relationship between MR and AR for a monopolist whose demand function is of the form P = a - bq

49 Worked Example 6.: Use differentiation to determine maximum TR Given the demand function P = 5 Q, (a) write down the equations for total revenue and marginal revenue (b) Calculate the output at which total revenue is a maximum and confirm that marginal revenue is zero at that point (a) Sketch the TR function and MR functions Confirm the answers in (b) graphically

50 Use differentiation to determine maximum TR where TR = 5Q - Q Step 1: Get derivatives (TR )ʹ = 5-4Q...equation (1) (TR )ˊˊ = equation () Step : Maximum ( or minimum!) when the first derivative is zero Hence, (TR)ʹ = in equation (1) and solve for Q = 5-4Q. Hence Q = 5/4 = 1.5 Step b: Substitute Q = 1.5 into TR = 5Q - Q Hence TR = 31.5 Step 3: In this case the second derivative is a constant, It is a negative constant, therefore this is a maximum point

51 (a) Sketch of TR = 5Q - Q : Note: This quadratic was graphed in Worked Example 4.7. TR TC 35 TR = Q = Q Figure 6.3 (part of.. See text for complete graph) Worked Example 6.1 :

52 Determine the output Q that maximises profit Determine maximum profit when given the equations for TR and TC, as follows: Method 1. Derive the equation for profit, Profit = Total Revenue -Total Cost, that is π = TR - TC then work through the max/min method to optimise π Method. Calculate the marginal revenue and marginal cost Then, profit is (a) maximized (b) minimized when MR = MC when MR = MC and (MR)ˊ < (MC)ˊ and (MR)ʹ > (MC)ʹ

53 An Example on profit Maximization When: TR = 5Q - Q and TC = 16 + Q (b) Graph the TR function and the TC function (c) Estimate the break-even points graphically, hence state the range of output, Q, for which the firm makes a profit. (d) From the graph estimate maximum profit (e) Use differentiation to determine maximum profit This problem is based on Worked Examples 6.1 and 6.

54 Graph TR = 5Q -Q and TC = 16 + Q See Worked Example 6.(a)and (b)(i) T C TR 35 3 Q = 1.5, Maximum TR = Profit TC = 16 + Q 15 1 Loss Loss Q Figure 6.4

55 (c) At break-even: TR = TC See Worked Example 6.(b)(ii) TR TC 35 Q = 1.5, Maximum TR = Profit TC = 16 + Q 15 Loss Loss 1 5 Break-even points Q = 4 and Q = Q Figure 6.4

56 (d) Maximum profit is difficult to determine graphically It is point where the difference: (TR TC) is greatest TR TC 35 Profit= TR - TC 3 5 TC = 16 + Q 15 Loss Loss 1 5 Break-even points Q = 4 and Q = Q Figure 6.4

57 (e) Use Differentiation to determine maximum profit See Worked Example 6.(c) First derive the equation for profit Profit = TR - TC = 5Q - Q - (16 + Q) π = 48Q - Q - 16 Now work through the max./ min. method to determine maximum profit Step 1: Get derivatives: πʹ = 48-4Q : πʹʹ = - 4 Step a: Slope, πʹ = at turning point, so, = 48-4Q giving Q = 1 Step b: When Q = 1, Profit, π = 48Q - Q - 16 = 48(1) - (1) - 16 = 18 Step 3: Second derivative = - 4, a negative constant, therefore profit is a maximum at Q = 1

58 Profit, π = TR - TC = 48Q - Q - 16 TR TC 35 3 TR = 31 Max. Revenue = 31.5 at Q = 1.5 Maximum Profit = 18 when Q = 1 Max. Profit = = 18 5 TC = 184 TC = 16 + Q 15 1 Q = 4 Q = Q Q = 1

59 Graph of profit function, confirming break-even points and maximum profit Maximum profit = 18 at Q = 1 Maximum profit = π = 48Q - Q Q = 1-1 Q Break-even points Figure 6.5: Profit function

60 Method : for Profit Maximization Given: TR = 5Q - Q and TC = 16 + Q Profit is maximized when (1) MR = MC and () (MR) ʹ < (MC)ʹ Start by deriving expressions for MR, MC, (MR)ʹ and (MC)ʹ MR = 5-4Q: MC = : (MR)ʹ = - 4 and (MC)ʹ = Apply condition (1) MR = MC: 5-4Q = 48 = 4Q Q = 1 So, a turning point ( a max or min) at Q = 1.

61 Method : Profit Maximization, continued TR = 5Q - Q and TC = 16 + Q Profit is maximized when (1) MR = MC and () (MR) ʹ< (MC)ʹ MR = 5-4Q: MC = : (MR)ʹ = - 4 and (MC)ʹ = From condition (1) MR = MC at Q = 1 So, there is a turning point at Q = 1. Now apply condition () to confirm that this is a maximum (MR)ʹ = - 4 and (MC)ʹ =. Hence, (MR)ʹ < (MC)ʹ Condition is true, confirming profit is a maximised. Maximum Profit = 18 when Q = 1

62 Graphs of TR, TC, MR and MC Maximum profit at the point where MR = MC π Maximum profit = 18 at Q = π = Q + 48 Q Q MR = 5-4Q: MR = MC MC = : MR = at Q = 1.5

63 Marginal and average functions Figure 6.9: P = a - bq: P = 5 - Q Note: the slope of the MR is twice the slope of the AR function P AR MR a = 5 MR = 5 4Q a / = 5 AR = 5- Q a/b = 5 Q a / b = 1.5

64 Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : SLIDES OF CHARTS. APPLICATIONS. John Wiley and Sons 13

65 Worked Example 6.5 Figure 6.7 Profit Maximization for perfect competition Worked Example 6.6 Figure 6.8 Profit Maximization for a monopolist Worked Example 6.31, Figure 6.37: Total Cost, Fixed Cost, Marginal Cost and Variable Cost Worked Example 6.3: Figure 6.38 TC, TVC, MC, AVC, AC Worked Example 6.3: Figure 6.38 (b) Points of intersection for MC, AC, AVC: Worked Example 6.9/6.3: Figure 6.36 Short run Production function and MP L, and APL: Slide 6 Worked Example 6.4, Figure 6.41: The relationship between TR, MR and elasticity of demand: Slide 7

66 Profit Maximization for perfect competition Worked Example (a) TC 4 TR Q -1 - Q = Q = 18 π (b) Profit maximization MC 4 3 Profit minimization MR 1 Q Figure 6.7

67 Profit Maximization for a monopolist Worked Example (a) TC 5 TR 15 1 Q = 15 5 Q = 1 Q π (b) Profit maximization MC 5 Profit minimization MR Q Figure

68 Total Cost, Fixed Cost, Marginal Cost and Variable Cost C (a) TC TVC 4 Points of inflection 3 1 Q (b) C Minimum MC = 5 MC Q Figure 6.37

69 Relationship between TC, MC, AVC, AC (a) Figure 6.38: Worked Example C 5 (a) TC, TVC, TFC functions TC 4 TVC 3 Points of inflexion 1 TFC Q C 14 (b) AC, AVC, MC functions Minimum AVC Minimum AC MC 4 Minimum MC AC AV C Q

70 Points of intersection for MC, AC, AVC Graph MC. AC. AVC as given in Worked Example C 1 (b) AC, AVC, MC functions Minimum MC Minimum AC Minimum AVC MC AC AVC Q Figure 6.38(b): Worked Example 6.3

71 Worked Example 6.9 / 6.3 (web) Point of inflexion on the production function Q (a) Short-run production function Q = f ( L ) 15 Point of inflexion 1 5 L MP L 3 APL 5 15 Maximum MP L Maximum APL APL 1 5 L (b) MP L, APL functions Figure 6.36 Short-run production function, MP L and APL functions

72 The relationship between TR, MR and elasticity of demand Figure 6.41: Worked Example 6.4 P MR MR = 6-1. Q (MR = a -bq) 4 P = Q 3 (P = a - bq) Q - -3 TR ε ε d d > 1 < 1 Q = a b ε ε 5 1 d d = 1 = 1 ε ε d d < 1 > 1 TR = 6 Q. 6 Q Q = a/b Q

73 Essential Mathematics for Economics and Business, 4 th Edition CHAPTER 6 : CURVATURE AND APPLICATIONS. John Wiley and Sons 13

74 Curvature: definitions: concave up and concave down: Slide, 3 Worked Example 6.7 (b): Figure Slide 4 Points of inflexion: definitions and diagrams: Slide 5, 6 Figure 6.34: Points of inflexion: Slide 7 Applications: Slide 8, 9: Short-run production function Figure 6.36 Applications: Slide 1, 11: Figure 6.37: Total cost and marginal cost

75 Curvature Concave up The curvature in the interval about a minimum where is described as concave up Concave down The curvature in the interval about a maximum where is described as concave down > y y < y > y <

76 Curvature y Concave up is sometimes described as convex towards the origin y Concave down is sometimes described as concave towards the origin y > y < x x

77 Curvature Worked Example 6.7 (b): Figure 6.33 Q Q = 5/L Q = 5 L d( Q) dl 5 = L L Hence Qʹʹ > d ( Q ) 5 = 3 dl L when L > Curve is convex towards the origin

78 Points of inflexion (PoI) The point of inflexion is the point at which curvature changes y < along the interval in which curvature is concave down y = at the point at which curvature changes (PoI) y > along the interval in which curvature is concave up y < y = y >

79 Points of inflexion The point of inflexion is the point at which curvature changes y < along the interval in which curvature is concave down y > along the interval in which curvature is concave up y = at the point at which curvature changes (PoI) y = y < y >

80 Points of Inflexion Figure 6.34 Points of inflexion at A1 and A You could say y is increasing at a increasing rate when yʹʹ > y < y > y = A y = A1 Points of inflexion y > y < You could say y is increasing at a decreasing rate when yʹʹ <

81 Applications of Points of inflexion Production functions and marginal products of labour Production Functions: See Figure 6.36 (web) MP L is increasing up to the PoI: MP L is decreasing after the PoI MP L is a maximum at the PoI: The value of L at which MP L is maximized is the value of L at which the point of inflection (at L = 1) occurs on the production function The PoI is described as The Law of diminishing returns to the factor labour :

82 Point of inflexion on the production function Q (a) Short-run production function Point of inflection Q = f L ( ) L MP L APL 3 5 Maximum MP L Maximum APL 15 AP L L (b) MP L, APL functions Figure 6.36 Short-run production function, MP L and APL functions

83 Applications of Points of inflexion The point of inflection on the normal total cost functions: (a) The marginal cost is decreasing before the PoI, then (b) increasing after the point of inflexion See Figure MC is minimized at the point of inflection (at Q = 1) on the total cost curve

84 Figure 6.37: Total cost and marginal cost 7 C 6 (a) TC 5 Points of inflection TVC C (b) Minimum MC MC Q Figure 6.37