Institut Perguruan Tun Hussein Onn

Coursework

Mathematics II

Name : Tan Chee Cung Lau Jin Yong Lee Yih Chuinn No.I/C : 910806-02-5841 910203-11-5119 910716-08-6579 Course: PPISMP Semester 3 Intake : July 2009 Option : MT / BI / BC Lecturer: Mr. Hamzah

Index

1.

Part B

a)

To find the dimension of the can that will minimize the cost, here is the method.

Given surface area, A = 2πr² + 2πrh, and πr²h=1.

Differentiate implicitly,

.∙. dAdr

= 4πr + 2πh + 2πrh’ and simplified by divided with 2π

.∙. dAdr

= 2r + h + rh’ … (1)

Since it is minimum, dAdr

= 0

1 = πr²h = V

.∙. dVdr

= 2πrh + πr²h’ and simplified by divided with πr

.∙. dVdr

= 2h + rh’ … (2)

Since we need minimum area, so dVdr

= 0

(2) – (1)

.∙. 2r-h = 0

.∙. hr

= 2

So the dimension of the metal can is h=2r. The height of the metal can must be

the diameter of the metal cylindrical can.

b)

These are the products we found in supermarket Pacific in Batu Pahat Mall. All

products are in cylindrical metal can. We measured it by using two rulers. We put one

ruler at top and one ruler at the can’s side, so that we can measure the height and the

diameter of the cans. There are some evidences and photos that we took from that

research.

Figure 1

This is the tea pot can that we took from that supermarket. We measured it and

we got the result as shown below:

d : 7.40cm

h : 10.30cm

where d stand for diameter, h stand for height.

Figure 2

This is another product that we took, that is F & N can. Although it is the same

can with tea pot can, but it consist of different d. This can also in cylindrical formed.

d : 7.10cm

h : 10.30cm

Figure 3

This Milo can is cylindrical formed and is a bit longer compare to products shown

before this. We took measured and below is the result we got.

d : 5.10cm

h : 13.00cm

Figure 4

Here is another type of cylindrical product we took from that supermarket, that is

Food Carrier. By neglecting the carrier, only that two containers will formed a cylindrical

shape. After we took the measurement, we got the result as shown below.

d : 16.20cm

h : 14.70cm

Figure 5

This is the last products we took from Pacific. Actually this mug contained a

cover, but we remove it because with the cover, it won’t be in cylindrical shape. After

removed that cover, we measured it. Result as shown below:

d : 10.50cm

h : 9.90cm

Tabulation of Data ( Measurements Taken)

Products Diameter, d Radius, r Height, hRatio,

hr

Tea Pot Can 7.40cm 3.70cm 10.30cm 10.303.70

= 10337

F & N Can 7.10cm 3.55cm 10.30cm 10.303.55

= 20671

Milo Can 5.10cm 2.55cm 13.00cm 13.002.55

= 26051

Food Carrier 16.20cm 8.10cm 14.70cm 14.708.10

= 4927

Mug 10.50cm 5.25cm 9.90cm 9.905.25

= 6635

From that table above shown, we can see different ratios. Now we are going to

calculate the average ratio of height to radius.

Average ratio = ( 10337 +

20671 +

26051 +

4927 +

6635 ) ÷ 5

= 2.8968

≈ 20069

c)

To prove that hr=8π

, here is the solution that can be used.

Surface area of the cylinder :

A = 8r² + 2πrh

A’ = 16r + 2πh + 2πrh’

Divided by 2:

A’ = 8r + πh + πrh’

Given the amount of waste metal is minimized:

A’ = 0

8r + πh + πrh’ = 0 ………………………………… Equation 1

We know that volume of the metal can is :

V = πr²h

h = Vπr ² ………………………………… Equation 2

Differentiate the equation:

V = πr²h

V’ = 2πrh + πr²h’

Divided by πr:

V’ = 2h + rh’

Given we need the V is minimized, so V’ = 0

V’ = 0

2h + rh’ = 0

rh' = -2h ………………………………… Equation 3

Substitute equation 3 into equation 1:

8r + πh + π(-2h) = 0 …………………………………….Equation 4

After that, substitute again the equation 2 into the equation 4:

8r + π ( Vπr ²

) + π (-2h) = 0

8r + Vr ²

-2πh = 0

Substitute h = Vπr ² again:

8r + Vr ²

-2π ( Vπr ²

) = 0

8r + Vr ²

-2 ( Vr ²

) = 0

Substitute V = πrh² into the equation:

8r + πr ²hr ²

-2 ( πr ²hr ²

) = 0

8r + πh – 2πh = 0

8r – πh = 0

8r = πh

hr

= 8π

= 2.546

≈ 2.55 (proved)