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The Misadventures of the Kacang Puteh Man Done by: Aisyah, Benedict, Ming Yi, Wen Qing, Zachary How we helped him discover the best container yet!

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The Misadventures of the Kacang Puteh Man Done by:Aisyah, Benedict, Ming Yi, Wen Qing, Zachary

How we helped him

discover the best

container yet!

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Brief IntroductionKacang Puteh is a local Singaporean/Malaysian snack which was once extremely popular in these 2 countries. Directly translated, it means “White Beans” and rightly so, as it comprises of a wide array of assorted nuts and snacks, making it a tasty snack. Most of the time, it would be served in a waxed paper cone or a plastic bag, each for differing purposes. Once very popular among students and moviegoers as a snack, it is now a dying art with the advent of popcorn and rising prices. Here are some of the ‘staples’ most commonly used in Kacang Puteh: Kacang Kuda (Chickpeas), assorted Murukku (an Indian snack), Kacang Parang (Broad Beans), Pagoda, Kacang Dhai, Dried Green Peas, and other assorted nuts and beans. Seeing the popularity of the cone shaped containers for easy eating, we decided to modify and improve on the most prominent shape used for Kacang Puteh containers. We decided that a square-based pyramid would be the best choice.

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Kacang Puteh Man

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Why did the Kacang Puteh Man use cones?

0 Substantial volumetric capabilities0 Quick and easy to make

(regardless whether made on the spot or not)0 Low material cost -- Any paper can be used.

Eg: Newspaper, Yellow Pages, School Exercise Book 0 Convenient storage

(cones stack into each other easily, saving plenty of storage space)

0 Easy to hold (tip of cone is slender)

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Kacang Puteh Container Designed By Group

Done by:Aisyah, Benedict, Ming Yi, Wen Qing, Zachary

A Step-by-Step

How-to!

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1. Get an A4 sized paper

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2. Fold the paper into 4 equal rectangles

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3. Draw a diagonal line across each quarter to get 3 isosceles triangles and 2 halves of an isosceles

triangle.

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4. Cut the black solid lines

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5. Rearrange the triangles to form a fan shape, with the 2 smaller triangles at opposite ends

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6. Tape along the black solid lines

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7. Fold the black solid lines to form a square base pyramid.

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8. Tape the sides together.

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9. Ta-da! Fill up with nuts/popcorn

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10. Fold along the middle and hold two different types of nuts. With the same total volume, the container is able to store two different types of

nuts/popcorn separately!

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11.There are many variations to this pyramid!

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12. There are many variations to this pyramid!

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Mathematical Calculations

Done by:Aisyah, Benedict, Ming Yi, Wen Qing, Zachary

A Step-by-Step

How-to!

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Necessary Information

Volume (V) = (1/3) b2h

Surface Area with open top (A) = 2bs

Dimensions of A4 sized paper: 210mm X 279mm

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Workings0 Since the entire piece of paper is used,

surface area (A) of the tetrahedral = surface area of paper.

0 2bs = 210 X 279

0 S = 29295/b -------------------- 1

Equations:S = 29295/b -------------------- 1

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Workings0By pythagoras theorem, 0h2 + (b/2)2 = s2

0h= √ (s2 – b2/4)0Sub 1 into equation:

0 h=√ [ (4(29295)2-b4)/4b2] --------- 2

Equations:

S = 29295/b -------------------- 1 h=√ [ (4(29295)2-b4)/4b2] --------- 2

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Workings

Since V = (1/3) b2h,

Sub 2 into V:

V= (b2/3)*√(4(29295)2- b4)/2b=b/6*√ ((29295)2- b4)

Equations:S = 29295/b -------------------- 1

h=√ [ (4(29295)2-b4)/4b2] --------- 2

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Workings

0Next, we differentiate V with respect to b:

0dV/db= d/db [(b/6)√((29295)2- b4)= -( b4-1144262700)/(2√(58590-b2)*√(b2+58590))

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Workings

0Then we find the value of b at the stationary point by using dV/db = 0:

When dV/db = 0,-( b4-1144262700)/(2*√ (58590-b2)* √ (b2+58590)) =

0-b4+1144262700 = 0Therefore,

b4= 1144262700b= 183.92 (2dp)

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Workings

0Next, to check if V is maximum at b = 183.92mm, we use second derivatives:

0d2v/db2

= √ (58590- b2)*√(b2+58590)*(x7-5721313500*x3)/(x8-6865576200* b4+11784034139501610000)

0 When b = 183.92,0d2v/db2 <00Therefore V is maximum at b = 183.92mm

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Workings0However, based on the dimensions of an A4 paper, b

cannot be more than 139.5mm

Therefore, we can only have a maximum of 139.5mm for b.

Subbing b=139.5mm into previous equations, we find that:

S = 210mm and h = 198.08mm (2dp)

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Conclusion

Optimum dimensions of a tetrahedral using an A4 piece of paper is:

b= 139.5mms= 210mmh= 198.08mm

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Contributions:0 Aisyah: Contributed to calculations and explanations of model. Collate

all materials. Uploaded information and answer onto blog. Helped discovered optimum shape of container.

 0 Benedict: Constructed actual Kacang Puteh container. Took photographs

of model created. Helped discovered optimum shape of container.

0 Ming Yi: Wrote brief write-up. Helped discovered optimum shape of container.

 0 Wen Qing: Worked-out mathematical concept of model. Calculated and

gave explanations on model. 0 Zachary: Wrote brief discussion on choice of packaging by the Kacang

Puteh man.