Endeavour College: Can Optimization Calculus Investigation, Stage 1

This project, completed by Rosie Nguyen Kavel East for Stage 1 Mathematical Methods at Endeavour College, investigates the optimization of cylindrical cans using calculus. The investigation explores minimizing the surface area of cans with fixed volumes (450mL, 500mL, and 550mL) to determine optimal radius, height, and height-to-radius ratios. The project involves setting up primary and secondary equations, determining derivatives, and analyzing the results to formulate a conjecture about the optimal can dimensions. The project also examines the efficiency of different packing methods (square, hexagonal, and triangular) and their impact on material waste. The analysis includes proofs, calculations, and comparisons to determine the most efficient packing method and optimal can dimensions. The conclusion highlights the optimal height-to-radius ratio and the most efficient packing method for minimizing material usage and production costs, supported by references to various mathematical and optimization resources.

Rosie Nguyen
Kavel East
CAN OPTIMISATION CALCULUS INVESTIGATION 2019
By (Firstname Lastname)
Mathematics
Name of Professor
Name of University
October 12, 2019

Rosie Nguyen
Kavel East
Stage 1 Mathematical Methods
CAN OPTIMISATION CALCULUS INVESTIGATION 2019
Introduction
The problem to be explore
To find the greatest or least value of a function using constraints, optimization is the way to
go. Using optimization, the correct and efficient value for a function of given domain. In
most disciplines, optimization is applied in physics, computer science, mathematics and
economics. In physics the process is applied in energy optimization. The optimal solution is
always expected to be at the first derivative function of an equation f ( x ) . Second derivative
test is used to proof whether the value minimum point obtained is with positive or negative.
(Gregory, 2018)
Method of solution to be used
The methods of optimization involve writing of primary equation of what to be optimized.
Dependent variable represents the quantity to be maximized with other independent variables.
Secondary equation is then determined and a relationship is stated for a variable with only
one independent variable (Mark, 2018). The feasible domain is written. First derivative is
determined to find the required maximum and minimum values. The answer to the problem is
established hence solving the actual question
Real-life applications of the topic
Application of optimization is used in the shaping of cylindrical cans. The manufacturing,
storage and transportation of these cans cost a lot of money and most companies have
invested in researching on how efficient they can produce a can that when used, will

Rosie Nguyen
Kavel East
minimize the production cost (Daniel , 2015). Cylinders were made by cutting a product
rectangle and wrapping it around with a single seam weld to which were added two circular
end caps. We have no edges in the hoop planes although round in part, thus reducing tension
levels when the is under strain. When you position them on a table, cylindrical cans remain in
place and are fairly difficult to open with a simple can opener. When contained in shells, they
pack more effectively than spheres, although not as effective as cuboids (DataGenetics, 2014)
Another application of the optimization in real life situation is packing of cylinders. The aim
of this approach is always to find the most efficient packing method that will lead to
optimization of the production cost. The application is based on packing cycles on a square
base, triangular base or circular base (Kim, et al., 2014). The results are evident for small
numbers of triangles, have an apparent symmetry, and are proved to be the best numerically.
Nonetheless, the best answers were not confirmed, and even known / revealed, over about
twenty. Specific closely resembling algorithms have been used to evaluate previously known
best outcomes (DataGenetics, 2019)
Methods Analysis and Assumptions
Part A:
Section 1
In this session, three different cans of fixed volumes (450mL, 500mL and 550mL) will be
investigated and the optimum radius, height and h
r ratio for each can will be determined.
The dependent variable will be is the quantity of the aluminium to be used. Therefore, we
will minimize the surface area of the cylinder
The surface area is given by A=2 π r2 +2 πrh

Rosie Nguyen
Kavel East
The surface is equation contain 2 variables i.e. radius r and height h. With the volumes given
we can make a secondary equation relating the variables and volume (Brown, 2014)
For volume V=400 ml
Volume is given by V =π r2 h=400
Since height appears once in the primary equation of determining surface area, we find h
from secondary equation h= 400
π r 2
We substitute h= 400
π r 2 ∈the surface areaequation
A=2 π r2 +2 πr 400
π r2 =2 π r2 + 800
r
We then determine the first derivative of the Surface area equation and set it to zero
dA
dr =4 πr− 800
r2 =0
4 πr = 800
r2
r3 =800
4 π
r ≈ 3.9929 cm
Replacing the value of radius r, in surface area equation we get h
h= 400
π∗3.99292 ≈ 7.9861 cm
the h: r ratio is equal to h
r = 7.9861
3.9929 =2.000075=2.0000
(Michelbrink, 2010)

Rosie Nguyen
Kavel East
For volume V=500 ml
Volume is given by V =π r2 h=500
Since height appears once in the primary equation of determining surface area, we find h
from secondary equation h=500
π r2
We substitute h=500
π r2 ∈the surface area equation
A=2 π r2 +2 πr 500
π r 2 =2 π r2 + 1000
r
We then determine the first derivative of the Surface area equation and set it to zero
dA
dr =4 πr− 1000
r2 =0
4 πr =1000
r2
r3 =1000
4 π
r ≈ 4.3013 cm
Replacing the value of radius r, in surface area equation we get h
h= 500
π∗4.30132 ≈ 8.6024 cm
the h: r ratio is equal to h
r = 8.6024
4.3013 =1.99995=2.0000
(Kamien & Schwartz, 2013)
For volume V=600 ml

Rosie Nguyen
Kavel East
Volume is given by V =π r2 h=600
Since height appears once in the primary equation of determining surface area, we find h
from secondary equation h=600
π r2
We substitute h=600
π r2 ∈the surface area equation
A=2 π r2 +2 πr 600
π r 2 =2 π r2 + 1200
r
We then determine the first derivative of the Surface area equation and set it to zero
dA
dr =4 πr− 1200
r2 =0
4 πr =1200
r2
r3 =1200
4 π
r ≈ 4.5708 cm
Replacing the value of radius r, in surface area equation we get h
h= 600
π∗4.57082 ≈ 9.1415cm
the h: r ratio is equal to h
r = 9.1415
4.5708 =1.9999781=2.0000
(Siyavula, 2018)
Volume Radius Height Ratio h
r
400mL r ≈ 3.9929 cm h ≈ 7.9861 cm 2.0000

Rosie Nguyen
Kavel East
500mL r ≈ 4.3013 cm h ≈ 8.6024 cm 2.0000
600mL r ≈ 4.5708 cm h ≈ 9.1415 cm 2.0000
Part A (2)
Proof of conjecture
Recall the volume of a cylinder V =π r2 h
h= V
π r2
we wilminimize the surface area givenby A=2 π r2 +2 πrh
A=2 π r2 + 2 πrV
π r2
A=2 π r2 + 2V
r
We then find the first derivative∧set¿ zero ¿ determine turning point
dA
dr =4 π r2− 2V
r2 =0
4 π r2 =2 V
r
this is therealtionship between volume∧radius of minimum surface area
V =2 π r3 remember volume was givenby V =2 π r 2 h
Conjecture: For any volume of a cylinder, the most efficient dimension is one in which the ratio
of the height, h to the radius r i.e. h
r is equal to 2.00. thus, efficient can is one in which the height
is twice the radius or height is equal to diameter i.e. h=2r

Rosie Nguyen
Kavel East
2 π r3 =2 π r2 h
therefore h=2 r
Part B (1)
‘square packing of circles’
a. Percentage of metal wasted
Consider the circles
Taking the dimension of the square to be 1 unit
Then area of the square = 1-unit square
For one circle, radius r = ½ units
Area of the circle ¿ π r2=π ( 1
2 )
2
= π
4
For 4 circles, radius = ¼ units
Area covered by 4 circles 4∗π r2=4∗π ( 1
4 )2
= 4∗π
16 = π
4
It can be observed that no matter the number of circles in a square packing, the unit square
area will always be π
4 which is 78.54 % packing achieved .

Rosie Nguyen
Kavel East
Therefore, the percentage of metal wasted ¿ 100 %−78.54 %=21.46 %
Part B (1)
b. Find the radius, height, and h
r ratio of a 400mL can which will optimise the
production cost when the circles are square packed
i) Rule of the area
Radius of the can =r
Area of the base shape ¿ 2 r∗2 r =4 r 2
Volume=A rea of base x height
V =4 r2∗h
V
4 r 2 =h
h= 400
4 r2
ii) Rule of the total area of the metal required
SA=2 x SA square base +SA side
SA(r )=2× 4 r2 +2 πrh
SA ( r ) =8 r2+ 2 πr x 400
4 r2 =8 r2 + 200 π
r
SA ’(r)=16 r− 200 π
r2
SA ’(r)=16 r 3−200 π
r2 =0

Rosie Nguyen
Kavel East
16 r3−200 π
r2 =0
16 r3−200 π =0
16 r3=200 π
r3 =200 π
16
r =3
√ 25 π
2
r = 3.399 cm
h= 400
4 r2
h=¿ 8.6556 cm
ratio for h
r =2.55
Types of packing methods
Hexagonal packing
We have the hexagonal packing which is more efficient. The packing takes about
π √3
6 of the square base
This represents 90.69 %

Rosie Nguyen
Kavel East
the perccentage of waste=100 %−90.69 %=9.31 %
For optimal radius we relate the efficiency for square packing and that of hexagonal packing
if 78.54 % corresspond ¿ 3.399 cm
90.69 %= 90.69∗3.399
78.54 =3.9248 cm
Height is given by h= 400
4 r2 = 400
4∗3.92482 =6.4918 cm
The ration of h
r is 6.4918
3.9248 =1.6540
(Barbu, 2012)
Triangular parking
The efficiency in triangular parking ¿ 60.96 %
The percentage of material wastage ¿ 100−60.96 %=39.04 %
Radius = 0.2543 units of the square dimension
For optimal radius we relate the efficiency for square packing and that of hexagonal packing
if 78.54 % corresspond ¿ 3.399 cm
90.69 %= 60.69∗3.399
78.54 =2.6265 cm
Height is given by h= 400
4 r2 = 400
4∗3.92482 =14.4959 cm
The ration of h
r is 14.4959
2.6265 =5.519
Part B3
Comparing the ratios determined above

Rosie Nguyen
Kavel East
The ration in part A is equal to 2.00 and the ratio for part B (1) is equal to 2.55 and part B(2)
is equal to 1.65 for hexagonal packing and 5.52 for triangular packing. this shows that for
efficiency to be achieved during manufacturing and packing of the cylindrical cans, the
height h, of the cylinder should always be double the radius or a ratio less than 2 (Smith,
2018).
Analysis
After analysing three different cans of varying volumes. It was found that given volume of a
cylinder with unknown values of radius and height, one can use optimization techniques to
find the dimensions. Secondly, the research has found that for efficient, economical and cost-
effective type of can, the height should be twice the radius thus the ratio of an efficient can
dimension is height to radius =2
Lastly from the research of various method of packing, it has been determined that hexagonal
packing style is the most efficient of all the other types of packing. This is because for
hexagonal packing, smaller area goes on wastes while bigger percentage is place in use.
Generally, derivation of surface area equation is the key to solving optimized conditions.
Assumptions
During investigation on the efficient dimension of the can, which will m minimize coat of
production, and the best packing method suitable for a company, the following parameters
were assumed.
i. It was assumed that all cans were closed on both ends and therefore when
computing surface area, the areas of both top and bottom circles were considered
ii. The thickness of the cans was very small and they were considered negligible and
could not affect the true volume and surface area of the can