2. Brief Introduction
Kacang 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.
4. Why did the Kacang Puteh
Man use cones?
0 Substantial volumetric capabilities
0 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)
15. 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!
19. Necessary Information
Volume (V) = (1/3) b2h
Surface Area with open top (A) = 2bs
Dimensions of A4 sized paper:
210mm X 279mm
20. Workings
0 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
21. Workings
0 By pythagoras theorem,
0 h2 + (b/2)2 = s2
0 h= √ (s2 – b2/4)
0 Sub 1 into equation:
0 h=√ [ (4(29295)2-b4)/4b2] ---------
2
Equations:
S = 29295/b -------------------- 1
h=√ [ (4(29295)2-b4)/4b2] --------- 2
22. 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
23. Workings
0 Next, we differentiate V with respect to b:
0 dV/db
= d/db [(b/6)√((29295)2- b4)
= -( b4-1144262700)/(2√(58590-b2)*√(b2+58590))
24. Workings
0 Then 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 = 0
Therefore,
b4= 1144262700
b= 183.92 (2dp)
25. Workings
0 Next, to check if V is maximum at b = 183.92mm, we use
second derivatives:
0 d2v/db2
= √ (58590- b2)*√(b2+58590)*(x7-5721313500*x3)/(x8-
6865576200* b4+11784034139501610000)
0 When b = 183.92,
0 d2v/db2 <0
0 Therefore V is maximum at b = 183.92mm
26. Workings
0 However, 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)
27. Conclusion
Optimum dimensions of a
tetrahedral using an A4 piece of
paper is:
b= 139.5mm
s= 210mm
h= 198.08mm
28. 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.
