MAT 271: Optimization Project Report: Calculus of Can Design

This report analyzes the optimization of can design using calculus principles. The project involves measuring and analyzing various cans, calculating their volumes and surface areas, and applying calculus to determine optimal dimensions for maximizing volume or minimizing surface area. The report includes photographs of the cans, detailed calculations involving derivatives, and comparisons of different can models. Graphs of volume and surface area versus radius are presented, and the report discusses why manufacturers may not always prioritize optimal can dimensions. The student explores how calculus can be used to improve can design and reduce manufacturing costs, providing a comprehensive analysis of the topic. The report also includes a discussion on why manufacturers may not always prioritize the most optimized can dimensions. The report is a comprehensive analysis of can design optimization using calculus.

MAT 271 Optimization Project
Discussion: Optimization – Calculus Report
Introduction: Cans are obviously used for variety of requirements they are mostly used for the sake drink consumption.
However still the actual shape and the size of the cans will depend on the end requirements. Also the organization
design interest and their opinions in general will shape the final shape and size of the can employed. Well though the
intention of the company may be to optimize or not, it is possible to optimize the shape of the can to take in the
greatest volume at the minimum cost of the product. Well considering the cost of the product based on the material
cost can be minimized if the can shape is able to manage to provide maximum volume for the minimum material
investment. The following part of the discussion will attempt to answer few mathematical questions related with this
topic.
 How well the can volumes can be optimized?
 Yes it is possible to optimize the volume of the can to be maximum for the given surface area. Or
alternatively it is possible to optimize the material surface area to be minimum for the given volume of
the can. In both these cases, the can dimensions can be subjected to calculus iterations and optimum
values can be found for the radius and height of the can cylinder shapes.
 Yes it is possible theoretically if the proportion of the radius and heights are changed; an unoptimized
cylinder material can be melt and reshaped to provide a different can with optimized dimensions of
volume. It is possible to reshape the can to enclose maximum volume by optimizing the dimensions.
 % = Actual volume/Optimized volume
(Possible to compute by changing the dimensions of the cans)
 Manufacturer can make this can in the chepest way, by using an optimized can for the given volume rates. For
the given volume the height and the radius of the can be optimized to minimize the total surface area of the can.
 Possible a manufacturer can of given volume can be optimized to employ less amount of material and
this inturn can minimize the cost of making the can.
Assumptions: All the cans are used with material of zero thickness.
Photoes of the cans:
Two coca cola cans of different heights and radius are collected their volume is same but surface area is different

Figure 1 Can-1
Figure 2 Can -2
The two figures above are the photographs of two different cans collected for the study. The volume of the two cans is
apparently same. The height and diameter of the two cans are different as follows,
Table 1 Given dimensions of the cans
S.N0 Can Height Can Radius Volume Surface area
1(Can 1) 12.1 Cm 3.1 cm 365.3cm3 235.56cm2
2(Can 2) 10cm 3.41 cm 365.3cm3 214.148cm2

For the given volume the minimum possible surface area can be computed using the principles of calculus.
Assuming V is the volume of the can, r is the radius of the can and h is the height of the can, the volume of the can be
computed by the expression V=π*r2*h ---------------- (1)
Further the surface area of the can be computed using the equation A= π*d*h + 2*π*r2---------------- (2)
H = V/π*r2
For the given value of V, the value of A is to be minimized; this is the objective of the first instance.
Differentiating equation 2 with respect to r,
A = (π*d*V)/.(π*r2) + 2*π*r2
dA/dr = -2V/r2 + 4*π*r -----------------------(3)
d2A/dr2 =4V/r3 + 4*π----------------------------- (4)
Condition for A to be maximum or minimum is that the equation 3 need to equated to zero, this will yield
, r is equal to cube root of V/2*π,
Further substituting this value in 4, d2A/dr2 is positive, which mean that the A will be minimum with the given value of r=
cube root of V/2*π.
For this value the relation between r and h if found it will be r=h/2
This is the condition for minimum surface area for the given volume.
So for the current case of Volume = 365.3 cm3, the relation between r and h can be computed.
V=365.3 = π*r2*h and h=2r or alternatively 365.3 = 2*π*r3
Further r is computed as, 3.87 cm and corresponding height (h) = 7.75cm, corresponding minimum surface area will be
188.1 cm2
So the best possible dimensions for minimizing the cost of the can making will be given by,
Table 2 Final minimum surface area can
S.N0 Can Height Can Radius Volume Surface area
1(model 1) 7.75cm 3.87cm 365.3cm3 188.1cm2
Alternatively for the given set of the cans, it is also possible to estimate the maximum volume possible.
From table 1, the actual maximum surface area is 235.56 cm2, for maximum volume this can be solved as follows,

For maximum volume, Given maximum A = 235.56cm2 = π* d2, which mean diameter d = 8.661 and corresponding radius
= 4.3305 cm – Maximum Volume for this = π*r2*h = 2*π*r3 = 510cc.
Similarly for given maximum A = 214.148 cm2 = π*d2, the mean diameter d = 8.817 and corresponding radius = 4.4085cm
Corresponding maximum volume = 538.611cc.
Table 3 optimized can volumes for the given SAs
S.No Surface
area
Radius for maximum volume Corresponding maximum volume
1(model 2) 235.56cm2 4.3305cm 510cc
2(model 3) 214.148cm2 4.4085cm 538.611cc
Analysis and resuts:
1) Comparing the surface areas of the model1, model 2, model 3, the degree of optimization is same in all the
cases, the cost of the material for unit volume is same. However model 1 provides minimum surface area for the
selected volume of the can.
2) Model 2 and Model 3s will provide maximum volumes for the given surface areas of the cans.
3) Optimization wise all the models 1,2,3 are optimized to the same degree, the cost of the material(Area) per unit
volume is same for all the three optimized models.
4) A) Radius vs Volume for can 1
B) Radius Vs volume for can 2
510cc(max)
365.3cc(Act)
Radius 3.1cm
Radius on x axis
Volume on y axis
Volume on y axis
538.611CC(max)
365.3cc(Act)

5) A) Radius Vs Surface area plots of the cans (Can 1)
Radius Vs Surface area plots for the can 2
6) Discussion: Manufacturers may not use a can with the maximized volume or mimized cost, as it may not the
ultimate objective. Typical requirements like special appearance, different outlook from the competitors,
standard sizes etc will determine the final sizes of the cans by the manufacturers. However in most of the
engineering applications optimization of the dimensions will work.
Report
You will turn in a written report (paragraph form) that includes the following:
 A photograph of each of your cans.
 The measured height and radius and the computed surface area and volume of each can.
 Clear explanations (show your work) of how you calculated the maximum possible volume for your cans (Part I)
and the minimum possible surface area (Part II) – this is calculus, so be sure to include your derivatives.
 Calculations showing how well the volume of each can is optimized.
 Compare the volumes of your cans and discuss which is most optimized.
 A graph of each can's volume versus radius –include a mark at the point on the graph for the maximum possible
volume and a mark at the point on the graph for the actual volume.
Radius 3.41cm
SA
235.56cm2(Act)
188.1cm2(min)
Radius 3.1cm
Radius 3.41cm
cm
214.2cm2(Act)
188.1 cm2(Min)

 A graph of each can’s surface area versus radius – include a mark at the minimum possible surface area and a
mark at the actual surface area.
Lastly, discuss why you think manufacturers do not use a can with maximized volume or minimized cost.
Please type your report in MS Word format. Photographs of your cans may be taken using any device and then inserted
into your report. Graphs may be done using Excel or any other graphing program (geogebra.org or desmos.com are
good).
ADVICE:
1. Please do all your work in centimeters, round to the nearest hundredth, and keep π in all calculations until the end.
2. Use units!
3. When measuring the radius of each can, you need to be as accurate as possible. You can get the most accurate
radius if you measure the circumference of the can (wrap a piece of paper or string around the can and mark the
circumference, and then measure the amount you have marked off) and then calculate the radius using C = 2πr. In
class you will have access to a tape measure.
Grading
Use the grade rubric provided to help guide you in setting up your final computer generated paper.
Content Total
Points
Explanation of can measurements, including photos. 10
Detailed calculations for maximum volume, including any derivatives and
derivative tests. 25
Detailed calculations for minimum surface area, including any derivatives and
derivative tests. 25
Computation showing how well each can is optimized, including a comparison of
which can is best optimized. 10
Graphs of volume/surface area versus radius, including noted locations of the
optimal volume/surface area and the actual volume/surface area. 10
Discussion of why cans are not optimized for volume/surface area. 10
Introduction & conclusion; overall presentation (formatting, clarity, grammar, etc.) 10
Project Grade 100