Lesson 24: Optimization
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The document discusses optimization problems in a calculus course, highlighting topics such as fixed perimeter rectangles with maximum area and techniques for understanding and solving optimization problems. It includes evaluations of teaching methods, student feedback, and the importance of conceptual understanding. Additionally, it provides examples and outlines methods for finding maxima and minima using closed interval and derivative tests.
![Solution (Continued)
Let its length be ℓ and its width be w. The objective function
is area A = ℓw.
This is a function of two variables, not one. But the perimeter
is fixed.
p − 2w
Since p = 2ℓ + 2w, we have ℓ = , so
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
Now we have A as a function of w alone (p is constant).
The natural domain of this function is [0, p/2] (we want to
make sure A(w) ≥ 0).
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-21-320.jpg)
![Solution (Concluded)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-22-320.jpg)
![Solution (Concluded)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the critical points, we find = p − 2w.
dw 2
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-23-320.jpg)
![Solution (Concluded)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-24-320.jpg)
![Solution (Concluded)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
Since this is the only critical point, it must be the maximum.
p
In this case ℓ = as well.
4
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-25-320.jpg)
![Solution (Concluded)
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
dA 1
To find the critical points, we find = p − 2w.
dw 2
The critical points are when
1 p
0= p − 2w =⇒ w =
2 4
Since this is the only critical point, it must be the maximum.
p
In this case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-26-320.jpg)
![Recall: The Closed Interval Method
See Section 4.1
To find the extreme values of a function f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points x where either f′ (x) = 0 or f is
not differentiable at x.
The points with the largest function value are the global
maximum points
The points with the smallest or most negative function value
are the global minimum points.
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-34-320.jpg)
![Recall: The First Derivative Test
See Section 4.3
Theorem (The First Derivative Test)
Let f be continuous on [a, b] and c a critical point of f in (a, b).
If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local
maximum.
If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local
minimum.
If f′ (x) has the same sign on (a, c) and (c, b), then c is not a
local extremum.
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-35-320.jpg)
![Recall: The Second Derivative Test
See Section 4.3
Theorem (The Second Derivative Test)
Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in
(a, b) with f′ (c) = 0.
If f′′ (c) < 0, then f(c) is a local maximum.
If f′′ (c) > 0, then f(c) is a local minimum.
If f′′ (c) = 0, the second derivative test is inconclusive (this does
not mean c is neither; we just don’t know yet).
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-36-320.jpg)
![Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-54-320.jpg)
![Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
dQ p
6. = p − 4w, which is zero when w = .
dw 4
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-55-320.jpg)
![Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
dQ p
6. = p − 4w, which is zero when w = .
dw 4
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-56-320.jpg)
![Solution
1. Everybody understand?
2. Draw a diagram.
3. Introduce notation: Length and width are ℓ and w. Length of
wire used is p.
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
dQ p
6. = p − 4w, which is zero when w = .
dw 4
Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the critical point is the absolute maximum.
. . . . . .](https://image.slidesharecdn.com/lesson24-optimizationi034slides-091125081048-phpapp02/85/Lesson-24-Optimization-57-320.jpg)
Lesson 24: Optimization
- 1. Section 4.5 Optimization Problems V63.0121.034, Calculus I November 23, 2009 Announcements Written HW (inc. WebAssign from 4.4) due Wednesday, Nov. 25 in class Quiz 5 next week on §§4.1–4.4, 4.7 . . . . . .
- 2. Thank you for the evaluations Comments and requests Too fast, not enough examples . . . . . .
- 3. Thank you for the evaluations Comments and requests Too fast, not enough examples Not enough time to do everything Class is not the only learning time (recitation and independent study) I try to balance . . . . . .
- 4. Thank you for the evaluations Comments and requests Too fast, not enough examples Not enough time to do everything Class is not the only learning time (recitation and independent study) I try to balance Too many proofs . . . . . .
- 5. Thank you for the evaluations Comments and requests Too fast, not enough examples Not enough time to do everything Class is not the only learning time (recitation and independent study) I try to balance Too many proofs In this course we care about concepts There will be conceptual problems on the exam Concepts are the keys to overcoming templated problems . . . . . .
- 6. A slide on slides Pro “Good use of powerpoint” “Use of slides removes uncertainty that could arise from the chalkboard” “helpful when they were posted online and I could refer back to them” . . . . . .
- 7. A slide on slides Pro “Good use of powerpoint” “Use of slides removes uncertainty that could arise from the chalkboard” “helpful when they were posted online and I could refer back to them” Con “I wish he would use the chalkboard more” “Powerpoint slides are not a good way to teach math” “I hate powerpoint” . . . . . .
- 8. A slide on slides Pro “Good use of powerpoint” “Use of slides removes uncertainty that could arise from the chalkboard” “helpful when they were posted online and I could refer back to them” Con “I wish he would use the chalkboard more” “Powerpoint slides are not a good way to teach math” “I hate powerpoint” Why I like them Board handwriting not an issue Easy to put online; notetaking is more than transcription Improvable—if you have suggestions I’m listening . . . . . .
- 9. A slide on WebAssign If you find a mistake, please let me know. We are dropping 5 lowest assignments (roughly two weeks’ worth) . . . . . .
- 10. Also requested, more: Tree stretches Music Dancing I’ll see what I can do! . . . . . .
- 11. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
- 12. Leading by Example Example What is the rectangle of fixed perimeter with maximum area? . . . . . .
- 13. Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. . . . . . . . .
- 14. Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. . . . ℓ . . . . . .
- 15. Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. . w . . . ℓ . . . . . .
- 16. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. . . . . . .
- 17. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is fixed. . . . . . .
- 18. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is fixed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , 2 . . . . . .
- 19. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is fixed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2 . . . . . .
- 20. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is fixed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2 Now we have A as a function of w alone (p is constant). . . . . . .
- 21. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is fixed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2 Now we have A as a function of w alone (p is constant). The natural domain of this function is [0, p/2] (we want to make sure A(w) ≥ 0). . . . . . .
- 22. Solution (Concluded) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. . . . . . .
- 23. Solution (Concluded) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 To find the critical points, we find = p − 2w. dw 2 . . . . . .
- 24. Solution (Concluded) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 To find the critical points, we find = p − 2w. dw 2 The critical points are when 1 p 0= p − 2w =⇒ w = 2 4 . . . . . .
- 25. Solution (Concluded) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 To find the critical points, we find = p − 2w. dw 2 The critical points are when 1 p 0= p − 2w =⇒ w = 2 4 Since this is the only critical point, it must be the maximum. p In this case ℓ = as well. 4 . . . . . .
- 26. Solution (Concluded) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 To find the critical points, we find = p − 2w. dw 2 The critical points are when 1 p 0= p − 2w =⇒ w = 2 4 Since this is the only critical point, it must be the maximum. p In this case ℓ = as well. 4 We have a square! The maximal area is A(p/4) = p2 /16. . . . . . .
- 27. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
- 28. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? . . . . . .
- 29. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. . . . . . .
- 30. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. . . . . . .
- 31. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols . . . . . .
- 32. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols 5. If Q is a function of more than one “decision variable”, use the given information to eliminate all but one of them. . . . . . .
- 33. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols 5. If Q is a function of more than one “decision variable”, use the given information to eliminate all but one of them. 6. Find the absolute maximum (or minimum, depending on the problem) of the function on its domain. . . . . . .
- 34. Recall: The Closed Interval Method See Section 4.1 To find the extreme values of a function f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the critical points x where either f′ (x) = 0 or f is not differentiable at x. The points with the largest function value are the global maximum points The points with the smallest or most negative function value are the global minimum points. . . . . . .
- 35. Recall: The First Derivative Test See Section 4.3 Theorem (The First Derivative Test) Let f be continuous on [a, b] and c a critical point of f in (a, b). If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local maximum. If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local minimum. If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. . . . . . .
- 36. Recall: The Second Derivative Test See Section 4.3 Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. If f′′ (c) = 0, the second derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). . . . . . .
- 37. Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequalities non-closed, non-closed, – gets global non-bounded non-bounded extrema intervals intervals automatically – only one – no need for derivative inequalities Con – only for closed – Uses – More derivatives bounded intervals inequalities – less conclusive – More work at than 1DT boundary than – more work at CIM boundary than CIM . . . . . .
- 38. Which to use when? The bottom line Use CIM if it applies: the domain is a closed, bounded interval If domain is not closed or not bounded, use 2DT if you like to take derivatives, or 1DT if you like to compare signs. . . . . . .
- 39. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
- 40. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . .
- 41. Solution 1. Everybody understand? . . . . . .
- 42. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . .
- 43. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Known: amount of fence used Unknown: area enclosed . . . . . .
- 44. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Known: amount of fence used Unknown: area enclosed Objective: maximize area Constraint: fixed fence length . . . . . .
- 45. Solution 1. Everybody understand? . . . . . .
- 46. Solution 1. Everybody understand? 2. Draw a diagram. . . . . . .
- 47. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . . . . . .
- 48. Solution 1. Everybody understand? 2. Draw a diagram. . . . . . .
- 49. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. . . . . . .
- 50. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . . . . . .
- 51. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . ℓ w . . . . . . . . . . .
- 52. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. . . . . . .
- 53. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. . . . . . .
- 54. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2] . . . . . .
- 55. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2] dQ p 6. = p − 4w, which is zero when w = . dw 4 . . . . . .
- 56. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2] dQ p 6. = p − 4w, which is zero when w = . dw 4 . . . . . .
- 57. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2] dQ p 6. = p − 4w, which is zero when w = . dw 4 Q(0) = Q(p/2) = 0, but (p) p p2 p2 Q =p· −2· = = 80, 000m2 4 4 16 8 so the critical point is the absolute maximum. . . . . . .
- 58. Your turn Example (The shortest fence) A 216m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of its sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? . . . . . .
- 59. Your turn Example (The shortest fence) A 216m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of its sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? Solution Let the length and width of the pea patch be ℓ and w. The amount of fence needed is f = 2ℓ + 3w. Since ℓw = A, a constant, we have A f(w) = 2 + 3w. w The domain is all positive numbers. . . . . . .
- 60. . . w . . . ℓ f = 2ℓ + 3w A = ℓw ≡ 216 . . . . . .
- 61. Solution (Continued) 2A We need to find the minimum value of f(w) = + 3w on w (0, ∞). . . . . . .
- 62. Solution (Continued) 2A We need to find the minimum value of f(w) = + 3w on w (0, ∞). We have df 2A =− 2 +3 dw w √ 2A which is zero when w = . 3 . . . . . .
- 63. Solution (Continued) 2A We need to find the minimum value of f(w) = + 3w on w (0, ∞). We have df 2A =− 2 +3 dw w √ 2A which is zero when w = . 3 Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the critical point is a minimum, in fact the global minimum. . . . . . .
- 64. Solution (Continued) 2A We need to find the minimum value of f(w) = + 3w on w (0, ∞). We have df 2A =− 2 +3 dw w √ 2A which is zero when w = . 3 Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the critical point is a minimum, in fact the global minimum. √ 2A So the area is minimized when w = = 12 and √ 3 A 3A ℓ= = = 18. The amount of fence needed is w 2 (√ ) √ √ 2A 3A 2A √ √ f = 2· +3 = 2 6A = 2 6 · 216 = 72m 3 2 3 . . . . . .
- 65. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . . . . . . .
- 66. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . Answer The dimensions are 4ft by 2ft. . . . . . .
- 67. Solution Let h and w be the height and width of the window. We have π π ( w )2 L = 2h + w + w A = wh + 2 2 2 If L is fixed to be 8 + 2π , we have 16 + 4π − 2w − π w h= , 4 so ( ) w π 1 π A = (16 + 4π − 2w − π w) + w2 = (π + 4)w − + w2 . 4 8 2 8 ( π ) So A′ = (π + 4)w − 1 + , which is zero when 4 π+4 w= = 4 ft. The dimensions are 4ft by 2ft. 1+ π2 . . . . . .
- 68. Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions . . . . . .