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Designing A Water Bottle

Designing A Water Bottle. W hy do we choose this topic? Students are not willing to bring their own water bottle. T hey always buy the water from the tuck shop . D o not reuse those bottles and just throw them away!  Not environment al friendly!. Objectives.

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Designing A Water Bottle

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  1. Designing A Water Bottle

  2. 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! Objectives

  3. Design a water bottle • 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.

  4. h A By considering a prism Fixed Fixed Volume of the prism = Base area (A)  Height (h) Fixed Is the total surface area fixed? No!

  5. Perimeter of the base h 2 Base area h Base area By considering a prism • 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

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

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

  8. Right-angled triangle • Suppose the area of the triangle is 100 cm 2

  9. (14. 1, 36.92828) Plot p against b, then we find out

  10. Conclusion: The perimeter is the smallest if b  h i.e. the right-angled isosceles triangle. Right-angled triangle • Suppose the area of the triangle is 100 cm 2

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

  12. Isosceles triangle

  13. (60, 45.59014) Plot P against the angle 

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

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

  16. Parallelogram

  17. Parallelogram (90 , 40)

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

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

  20. Rectangle

  21. Rectangle ( 10, 40)

  22. Rectangle

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

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

  25. Consider the area of each n-sided polygon is fixed, for example,100cm2.

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

  27. Conclusion of the base We know that when the number of sides Its perimeter We decided to choose CIRCLEas the base of our water bottle.

  28. Sphere Cylinder Cone The Bottle In Form 3, we have learned the solid related circle, they are…

  29. Cylinder Although the volume of the cylinder is fixed, their total surface area are different. 1.5 cm 2.5 cm 3.5 cm 56.6 cm 10.4 cm 20.4 cm Volume = 400 cm3 Total surface area = 359.3 cm2 Volume = 400 cm3 Total surface area = 305.5 cm2 Volume = 400 cm3 Total surface area = 547.5 cm2

  30. h r Cylinder Cylinder • Suppose the volume of cylinder is fixed (400cm3), we would like to find the ratio of radius to height so that the surface area is the smallest. Volume = r2h Total surface area = 2r2 + 2rh

  31. Cylinder with cover

  32. (0.5, 300.531) Plot A against (r/h)

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

  34. L h r Cone Cone Suppose the volume of cone is fixed (400cm3), we would like to find the ratio of radius to height so that the surface area is the smallest. Volume = r2h Total surface area = r2 + rL

  35. Cone

  36. (0.353, 609.29) Plot total surface area against (r/h)

  37. h r Conclusion of the cone • Suppose the volume of the cone is fixed, the surface are is the smallest if

  38. If r : h = 1: 2 Cylinder Cone Volume = r2h = 2r3 = 6r2 Comparison r h = 2r h = 0.354r r r Sphere If r : h = 1: 0.353 Volume = r3 Volume = r2h Surface area = 4r2 = r3 Surface area = 2rh + 2r2 Surface area = r =2.06r2

  39. If the volume of the 3 solids are fixed, we would like to compare their total surface areas

  40. Compare their surface areas if their volumes are equal

  41. 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…….

  42. 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 = r2 + 2rh h r

  43. Cylinder without cover

  44. Find r : h of cylinder without cover (1.005 ,238.529)

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

  46. Conclusion From the graph, if r : h = 1: 1, the smallest surface area of cylinder will be attained. Volume of bottle = r2(r)+ r3 = r3 E.g. If the volume of water is 500 cm3, then the radius of the bottle should be 4.57 cm

  47. 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?

  48. 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

  49. References : Book: Chan ,Leung, Kwok (2001), New Trend Mathematics S3B, Chung Tai Education Press Website: http://mathworld.wolfram.com/topics/Geometry.html http://en.wikipedia.org/wiki/Cone http://en.wikipedia.org/wiki/Sphere http://www.geom.uiuc.edu/

  50. END END END

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