designing a water bottle

W hy 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!

• 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.

Volume of the prism = Base area (A) Height (h)

Fixed

Fixed

Fixed

A

hIs the total surface area fixed?No!

h

Perimeter of the base

h

Base area

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

First, we need to choose the base for our bottle. We start from the

basic figures.

ParallelogramTriangle

Part I: TrianglePart I: TriangleFirst, we begin with a right-angled triangle and assume the area is fixed.

b

h c2

1

b

A2Area (A) = bh h=

Perimeter (P) = b + h + c

= b + h +

= b + +

22 bh

b

A2 22

2

4b

b

A

Right-angled Right-angled triangletriangle• Suppose the area of the

triangle is 100 cm 2

Base(b)

Perimeter (p)

1 401.0025

2 202.02

: :

14.1 36.92828

: :

Plot p against b, then we find out

perimeter against base if base area is 100 cm square

0

50

100

150

200

250

300

350

400

450

0 20 40 60 80 100 120

base

Perim

eter

perimeter

(14. 1, 36.92828)

Right-angled Right-angled triangletriangle

• Suppose the area of the triangle is 100 cm 2

Area(A)

Base(b) Height(h) Hypotenuse(a)

Perimeter (p)

100 1 200 200.0025 401.0025

100 2 100 100.02 202.02

: : : : :

100 14.1 14.1844 20.00204 36.92828

: : : : :

Conclusion:The perimeter is the smallest if b h

i.e. the right-angled isosceles triangle.

Next, consider isosceles Next, consider isosceles triangletriangle

Base (b)

Height (h)

Length (l)

Consider 0o 90o

Area = base (b) Height (h)

= (2l cos) (l sin)

2

1

sincos

Al

Perimeter (p) = 2l + b

)cos22(sincos

)cos2(2

A

llp

2

1

Isosceles triangleIsosceles triangleangle in

degreeperimete

r (p)

1302.788

1

2214.119

8: :

6045.5901

4: :

Plot P against the angle Plot perimetre against the angle

0

50

100

150

200

250

300

350

0 10 20 30 40 50 60 70 80 90 100

Angle

Perim

etre

perimeter

(60, 45.59014)

Result from the graphFrom the graph, we know that the

perimeter is the smallest when = 60o

Each angle is 60o ( sum of )

Equilateral triangle has the

smallest perimeter

ParallelogramParallelogram

Height (h)

Base (b)

Side (l)

(where 0o 90o)

Perimeter (p) = 2(b + l)

= 2( + l)

Area (A) = b h (where

A is fixed) and h= l sin

b = sinl

A

h

A

sinl

A

angle perimeter

5 249.4798

10 135.1782

: :

60 51.11511

: :

90 40

Parallelogram

Perimeter

0

50

100

150

200

250

300

0 50 100

Perimeter

Parallelogram

(90 , 40)

From the graph,

• The perimeter is the smallest if = 90o

Rectangle gives

the smallest perimet

er

Result from the graph

Length (l)

Width (w)

Area (A) = Length (l) Width(w)

(where A is fixed)

l

Aw

Perimeter (p) = 2(l + w)

=2 (l + ) l

A

Rectangle

lengthperimete

r0.5 4011 202． ．． ．10 40． ．

99.5 201.01100 202

Rectangle

05101520253035404550556065707580859095

100

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100

length

perim

eter

( 10, 40)

Rectangle

Area(A) length widthperimeter

100 0.5 200 401

100 1 100 202

． ． ．． ． ．

100 10 10 40

． ． ．

100 99.51.005

03201.0

1

100 100 1 202

Rectangle

From the graph, if length = width, the rectangle has the smallest

perimeter.

Square gives the smallest perimeter

Result from the graph

PolygonPolygonFrom the above, we find out that regular figures have the smallest perimeter. So we tried out more regular polygons,

eg. ……

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

number of sides (n)

perimeter (p)

4 40

637.22419436

: :

3035.5140933

: :

Plot p against n if area is fixed

perimeter against number of side

35

3637

38

3940

41

0 20 40 60

number of side

perim

eter

perimeter

The perimeter is decreasing as the number of sides is increasing.

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

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

Cylinder Cone Sphere

Cylinder

1.5 cm2.5 cm

3.5 cm

56.6 cm20.4 cm 10.4 cm

Volume = 400 cm3

Total surface area

= 547.5 cm2

Volume = 400 cm3

Total surface area

= 359.3 cm2

Volume = 400 cm3

Total surface area

= 305.5 cm2

Although the volume of the cylinder is fixed, their total surface area are different.

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 = 2r2 +

2rhCylinder

r

h

Cylinder with coverArea (A) r/h806.2832 0.007854

734.8754 0.010454

675.7145 0.013572

626.0032 0.017255

583.7436 0.021551

547.4705 0.026507

516.085 0.03217

488.7466 0.038587

464.802 0.045804

443.7349 0.05387

Plot A against (r/h)

Plot surface area against (r/h)

0

100

200

300

400

500

600

700

800

900

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

r/h

Surface area r/h

(0.5, 300.531)

Conclusion of the cylinder

• Suppose the volume of the cylinder is fixed, the surface are is the smallest if

Cylinder

r

h

2

1

h

r

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

3

1r

Cone

hL

ConeConesurface area r/h610.990626 0.404648

610.785371 0.401299

610.594731 0.397996

610.418393 0.394738

610.256057 0.391525

610.107426 0.388354

609.972212 0.385227

609.850136 0.382141

609.740922 0.379096

609.644304 0.376091

609.560021 0.373126

Area against (r/h)

609

609.5

610

610.5

611

611.5

612

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

r/h

Surf

ace

area

r/h

Plot total surface area against (r/h)

(0.353, 609.29)

Conclusion of the cone

• Suppose the volume of the cone is fixed, the surface are is the smallest if

r

h

353.0h

r

Comparison

Cylinder Cone Sphere

If r : h = 1: 0.353

Volume = r2h

Surface area

= 2rh + 2r2

Volume = r2h

If r : h = 1: 2

= 2r3

= 6r2

3

1

= r3 3

353.0

Surface area

= r 22 hr =2.06r2

Volume = r3 3

4

Surface area = 4r2

r r

h = 2r h = 0.354r r

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

areasvolume

Surface(cylinder)

surface(cone)

surface(sphere)

100 119.2654 270.2723782 104.1879416

200 189.3221 429.0306575 165.3880481

300 248.0821 562.1892018 216.7196518

: : : :

1000 553.581 1254.493253 483.5975862

1100 589.8972 1336.790816 515.3226696

: : : :

Compare their surface areas if their volumes are equal

Compare their total surface areas if their volumes equal

0

500

1000

1500

2000

2500

0 500 1000 1500 2000 2500

Surface area

Vol

ume surface(cylinder)

surface(cone)

surface(sphere)

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…….

Our choiceThe designed bottle isCylinder + Hemisphere

In the case the cylinder doesnot have a cover. Therefore,we need to find the ratio ofradius to height of an opencylinder such that itssurface area is the smallest.i.e. Total surface area = r2 +

2rh

r

h

Cylinder without coverarea r/h

803.1415 0.007854

731.0739 0.010453

671.1904 0.013571

620.6938 0.017255

577.5859 0.021551

540.4017 0.026506

508.0422 0.032169

479.6672 0.038585

454.6229 0.045803

432.3934 0.053869

412.566 0.06283

Plot (A) against (r/h)

0

100

200

300

400

500

600

700

800

900

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

r/h

tota

l sur

face

are

a(A

)

r/h

Find r : h of cylinder without cover

(1.005 ,238.529)

Cylinder without cover• Suppose the

volume of the cylinder is fixed, the surface are is the smallest if

Cylinder

r

h

1

1

h

r

ConclusionFrom the graph, if r : h = 1: 1, the smallestsurface area of cylinder will be attained.

Volume of bottle = r2(r) + r3

= r3

3

4

2

1

3

5

E.g. If the volume of water is 500 cm3, then the radius of the bottle should be 4.57 cm

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?

Member listSchool : 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

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/

Natural Examples

watermelons

oranges

cherry

calabash

Chinese Design

bowl Wine container

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.

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.

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.