Differentiation
Dimensions of a Beverage Can
Presented by:
Tan Chee Meng
Ahmad Tajuddin
Our role

We are production managers in a
beverage company. Our task is to
determine the dimension of a can that is
cost effective and satisfies our customer’s
needs.
Our Goals





To determine the appropriate dimension of a
cylindrical beverage can
To find the radius and the height of the
cylindrical can when its total surface area is
minimum
To find the minimum surface area of the can
using various strategies
To find the relationship between the height
and radius of the can
To investigate the preference of customer in
terms of volume, dimension and aesthetic
value
Our Plan
Explore & Compare Various Strategies

Strategy 1: Apply differential calculus
 Choose suitable symbols to represent the variables:
radius (r), height (h), surface area (A) and volume (V)
 Formulate an equation of surface area (A) in terms of
radius
 Differentiate A with respect to r
 Find the turning point when dA/dr =0
 Substitute value of r to find A and h
 Repeat the above steps with different values of the
volume (V)
 Find the relationship between r and h
Our Plan
Explore & Compare Various Strategies

Strategy 2: Use Geometer’s Sketchpad





Plot the graph of f(r)
Find the first derivative of f(r) i.e. f’(r)
Plot the graph of f’(r)
Find the intersection of the graph f’(r) with the x-axis
Find the value of r (x-coordinate) when surface area (A)
is minimum
Our Plan
Explore & Compare Various Strategies

Strategy 3: Make tables using spreadsheet
 Make a table to find the surface area of the can with
different values of r and h (write formula to enable
Spreadsheet to calculate the required values
automatically)
 Make tables to show the value of A with different volume
of the can e.g. V=400cm3, 375cm3 etc.
Make a survey to determine the
preference and needs of the customers:
Collect data regarding the preferences and
needs of the customer in terms of the can
dimension, the volume and the appearance
(aesthetic value) through survey and Internet
research
Implement the Strategy
Find
general
equation
relating
variables
Formulate
equation
of A in
terms of r
where
V is a
constant
Find the
value of r
When V= 400
Find the values
of r , h and
minimum
surface area A
3
cm
When V= 375
3
cm
The values of r and h based on different
values of V are as follows:
Relationship between h and r (or D):
h = 2r= Diameter
The height of the can is
approximately equal to its
base diameter when the
surface area is minimum
The table below shows the surface area
of the can, its base radius and height
Difference in
surface area
12.7 cm2
12.9 cm2
13.3 cm2
13.6 cm2
The surface area is reduced from 12.7
to 13.6 cm2 when the radius is
decreased by 0.1 cm from 4 cm to
3.6cm
Graphical Method:
Using Geometer’s Sketchpad
When V= 400 cm3
f(r)
Minimum
value of
surface area
of can,
A=300.48 cm3
(3.99,300.48)
The function is
minimum if
f’(r)=0.
Find the
coordinates of
intersection
between the
graph f’(r) and
the x axis.
r
(3.99,0.00)
Graph 1
Value of r when
surface area is
minimum
f(r)=0
When V= 375 cm3
f(r)
Minimum
value of
surface area
of can,
A=287.83 cm3
(3.91, 287.83)
Find the
coordinates
of
intersection
between the
graph f’(r)
and the x
axis
(3.91, 0.00)
Graph 2
r
Plot the graph of h vs r to find the values of h as r varies
h
Move the point along
the line to determine
values of h and r
(coordinates)
Values
of r
Values
of h
r
Algebra method (differential) VS Graphical method (GSP)
ALGEBRA METHOD
(DIFFERENTIAL)
• Need to carry out
tedious calculations to
determine each value
of h and A
• Need to use scientific
calculator to calculate
the values
GRAPHICAL METHOD
(GSP)
• Can use GSP to plot
complicated graph of
function, its 1st and 2nd
derivatives.
• Able to determine the
minimum / maximum
value from the graph
with ease
• From the graph of
function relating h and
r, we can determine
the value of h for any
value of r by moving
the point along the
graph (determine the
coordinates)
Find the surface area using spreadsheet
When V=
400 cm3
(cm2)
Minimum
surface
area
=310.8
cm2
When
radius
=3.3 cm
Comparing with
the minimum
surface (300.5
cm3 )as r changes
Percentage
increase in the
surface area as
compared to
the minimum
surface area of
300.5 cm2
Minimum
surface area
=300.5 cm2
When radius
=4.0 cm
Table 1
When V=
375 cm3
(cm2)
When the radius
decreases, the total
surface area of the
can increases
significantly from 0.1
% to 17.9 %
Minimum
surface area
=287.8 cm2
When radius
=3.9 cm
Table 2
Data Analysis
Table 3 shows the preferred choice of the volume of
the drink in the can. Total number of people
participated in the survey is 128.
Table 3
Data indicate that the preferred volumes of drink in
the can are 325cm3 and 350 cm3 among customers.
We narrow down the choice of volumes of drink to
350cm3 or 325cm3.
Data Analysis
Table 4 shows the preferred dimensions of the can of
volume 350 cm3
Table 4
Data indicate that the preferred choice of diameter
and height of the can is 6.6 cm and 10.2 cm
respectively.
The elevation of the cylindrical can is a rectangle of
sides 6.6 cm x 10.2 cm. The ratio 10.2/6.6 = 1.55 is
very close to golden ratio which is aesthetically
pleasing to the eye.
Data Analysis
Table 5 shows the preferred dimensions of the can
of volume 325 cm3
Table 5
Data indicate that the preferred choice of diameter
and height of the can is 6.6 cm and 9.5 cm
respectively.
Potential Customers' Comments

“ The shape when r=6.6cm and 5 cm is
pleasing to the eye”

“It looks ugly if the height of the can is
much bigger than the diameter of the can.
Even though the surface area is minimum
when diameter equals height of can, the
side elevation of the can is a square. This
shape is not interesting”
Potential Customers' Comments

“ The can is nice to hold when its diameter
is 6.6cm. It is difficult to have a good grip
of the can if the diameter is too large”

“For a carbonated drink, about 325 cm3
will be sufficient to quench my thirst.
Discussions

The preferred choice of dimension for the
diameter of the can is 6.6cm with a height of
10.2 cm (for Volume of 350cm3) and 9.5 cm
(for volume of 325cm3).

We need to do a comparison between the
surface areas of the can of volume 350 cm3
and 325 cm3 as shown in table 6 so as to
choose the most suitable dimension that
reduces the surface area and hence the cost
of aluminum to make the can.
Compare the Surface Area
Difference
between
surface areas
V=350cm3 and
V=325cm3
When
diameter
of the
can is
6.6 cm
Percentage
difference
between
surface areas
V=350cm3 and
V=325cm3
5.4%
Table 6
Discussions

From table 6, when the diameter of the can is
6.6cm, the surface area of the can is
decreased by 5.4 % when its volume
decreases from 350 cm3 to 325 cm3. The
percentage of reduction is quite significant

Assume for a beverage company, the total
cost for the production of the cans is RM5
million, then reduction in cost will amount to
RM 270,000 i.e. (5.4/100) x 5 000 000

What will be the amount if the cost is RM 100
million ??
Current News
Think about its Implications..?
Think…

Can we use 3-D shape such as cuboid, sphere or
prism for packaging carbonated soft drinks?
Why?

Can we recycle the aluminum cans? What is the
cost of recycling the cans?

Can we use other cheaper materials other than
aluminum?

What is the actual dimensions of the beverage
can available at the local supermarket?

What is the possible future design and
dimensions of the can?
Conclusion
After considering customers’ perception
and needs, cost of production and the
rising price of aluminum and theoretical
calculation, we decided that the
dimensions of the can are as follows:


Volume of can = 325 cm3

Radius = 6.6 cm

Height =9.5 cm