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Why do we choose this topic? Students are not willing to bring their own water bottle. They always buy the water from the tuck shop. Do not reuse those.

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Presentation on theme: "Why do we choose this topic? Students are not willing to bring their own water bottle. They always buy the water from the tuck shop. Do not reuse those."— Presentation transcript:

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3 Why do we choose this topic? Students are not willing to bring their own water bottle. They always buy the water from the tuck shop. Do not reuse those bottles and just throw them away!  Not environmental friendly!

4 If the volume (V) of the bottle is fixed, we would like to design a water bottle so that its material used (total surface area) is the smallest.

5 Volume of the prism = Base area (A)  Height (h ) Fixed A h Is the total surface area fixed? No!

6 h Perimeter of the base h Base area 2 Total surface area = 2  Base area + total areas of lateral faces = 2  Base area + perimeter of the base  height Conclusion: The smaller the perimeter of the base is, the smaller is the total surface area

7 First, we need to choose the base for our bottle. We start from the basic figures. Parallelogram Triangle

8 Part I: Triangle First, we begin with a right- angled triangle and assume the area is fixed. b h c Area (A) = bh  h= Perimeter (P) = b + h + c = b + h + = b + +

9 Right-angled triangle Suppose the area of the triangle is 100 cm 2 Base(b)Perimeter (p) :: ::

10 Plot p against b, then we find out (14. 1, )

11 Right-angled triangle Suppose the area of the triangle is 100 cm 2 Area(A)Base(b)Height(h)Hypotenuse(a)Perimeter (p) ::::: ::::: Conclusion: The perimeter is the smallest if b  h i.e. the right-angled isosceles triangle.

12 Next, consider isosceles triangle  Base (b) Height (h) Length (l) Consider 0 o    90 o Area = base (b)  Height (h) = (2l cos  )  (l sin  )  Perimeter (p) = 2l + b 2 1

13 Isosceles triangle angle  in degree perimeter (p) :: ::

14 Plot P against the angle  (60, )

15 Result from the graph From the graph, we know that the perimeter is the smallest when  = 60 o  Each angle is 60 o (  sum of  ) Equilateral triangle has the smallest perimeter

16 Parallelogram Height (h)  Base (b) Side (l) (where 0 o    90 o ) Perimeter (p) = 2(b + l) = 2( + l ) Area (A) = b  h (where A is fixed) and h= l sin  b =

17 angle  perimeter :: :: 9040 Parallelogram

18 (90, 40)

19 From the graph, The perimeter is the smallest if  = 90 o Rectangle gives the smallest perimeter Result from the graph

20 Length (l) Width (w) Area (A) = Length (l)  Width(w)  (where A is fixed) Perimeter (p) = 2(l + w) =2 (l + ) Rectangle

21 lengthperimeter .. .. 1040 .. Rectangle

22 ( 10, 40) Rectangle

23 Area(A)lengthwidth perimet er ... ... ... Rectangle

24 From the graph, if length = width, the rectangle has the smallest perimeter. Square gives the smallest perimeter Result from the graph

25 Polygon From the above, we find out that regular figures have the smallest perimeter. So we tried out more regular polygons, eg. ……

26 Consider the area of each n-sided polygon is fixed, for example,100cm 2. number of sides (n) perimete r (p) :: ::

27 Plot p against n if area is fixed The perimeter is decreasing as the number of sides is increasing.

28 Conclusion of the base We know that when the number of sides We decided to choose CIRCLE as the base of our water bottle. Its perimeter

29 In Form 3, we have learned the solid related circle, they are… CylinderCone Sphere

30 Cylinder 1.5 cm 2.5 cm 3.5 cm 56.6 cm 20.4 cm 10.4 cm Volume = 400 cm 3 Total surface area = cm 2 Volume = 400 cm 3 Total surface area = cm 2 Volume = 400 cm 3 Total surface area = cm 2 Although the volume of the cylinder is fixed, their total surface area are different.

31 Cylinder Suppose the volume of cylinder is fixed (400cm 3 ), we would like to find the ratio of radius to height so that the surface area is the smallest. Volume =  r 2 h Total surface area = 2  r  rh Cylinder r h

32 Cylinder with cover Area (A)r/h

33 Plot A against (r/h) (0.5, )

34 Conclusion of the cylinder Suppose the volume of the cylinder is fixed, the surface are is the smallest if Cylinder r h

35 Cone Suppose the volume of cone is fixed (400cm 3 ), we would like to find the ratio of radius to height so that the surface area is the smallest. Volume =  r 2 h Total surface area =  r 2 +  rL r Cone h L

36 Cone surface arear/h

37 Plot total surface area against (r/h) (0.353, )

38 Conclusion of the cone Suppose the volume of the cone is fixed, the surface are is the smallest if r h

39 Comparison Cylinder Cone Sphere If r : h = 1: Volume =  r 2 h Surface area = 2  rh + 2  r 2 Volume =  r 2 h If r : h = 1: 2 = 2  r 3 = 6  r 2 =  r 3 Surface area =  r =2.06  r 2 Volume =  r 3 Surface area = 4  r 2 rr h = 2rh = 0.354r r

40 If the volume of the 3 solids are fixed, we would like to compare their total surface areas volume Surface (cylinder) surface(cone)surface(sphere) :::: ::::

41 Compare their surface areas if their volumes are equal

42 Conclusion From the graph, if the volume is fixed Surface area of sphere < cylinder < cone We know that sphere gives the smallest total surface area. However…….

43 Our choice The designed bottle is Cylinder + Hemisphere In the case the cylinder does not have a cover. Therefore, we need to find the ratio of radius to height of an open cylinder such that its surface area is the smallest. i.e. Total surface area =  r  rh r h

44 Cylinder without cover arear/h

45 Find r : h of cylinder without cover (1.005, )

46 Cylinder without cover Suppose the volume of the cylinder is fixed, the surface are is the smallest if Cylinder r h

47 Conclusion From the graph, if r : h = 1 : 1, the smallest surface area of cylinder will be attained. Volume of bottle =  r 2 (r) +  r 3 =  r 3 E.g. If the volume of water is 500 cm 3, then the radius of the bottle should be 4.57 cm

48 Open-ended Question Can you think of other solids in our daily life / natural environment that have the largest volume but the smallest total surface area?

49 Member list School : Hong Kong Chinese Women’s Club College Supervisor : Miss Lee Wing Har Group leader: Lo Tin Yau, Geoffrey 3B37 Members: Kwong Ka Man, Mandy 3B09 Lee Tin Wai, Sophia 3B13 Tam Ying Ying, Vivian 3B21 Cheung Ching Yin, Mark 3B29 Lai Cheuk Hay, Hayward 3B36

50 References : Book: Chan,Leung, Kwok (2001), New Trend Mathematics S3B, Chung Tai Education Press Website: ml

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52 Natural Examples watermelons oranges cherry calabash

53 Chinese Design bowl Wine container

54 Reflection: After doing the project, we have learnt : 1.more about geometric skills, calculating skills of different prisms, such as cylinders, cones and spheres 2.information research and presentation skills 3.plotting graphs by using Microsoft Excel 4.The most important thing: we learnt that we can use mathematics to explain a lot of things in our daily lives.

55 Reflection: Although we faced a lot of difficulties during our project, we never gave up and finally overcame all of them. We widened our horizons and explored mathematics in different aspects in an interesting way. Also, Miss Lee helped us a lot to solve the difficulties. We would like to express our gratitude and sincere thanks to her.

56 Limitation Our Maths knowledge is very limited, we wanted to calculate other designs like the calabash or a sphere with a flattened base, but it was to difficult for our level. Our knowledge in using Microsoft Excel has caused us a lot of technical problems and difficulties.


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