Calculus I (Sem1)
Project 1: Finding an Optimal Can
Dr M. Campbell
INTRODUCTION
Previously, we used the derivative
to find a can which had a given
volume with minimum surface
area.
A technique called the “calculus of
variations” can be used to show
the shape with minimum surface
area that holds a given volume is a
sphere . But it wouldn’t be
practical to store food in a
spherically-shaped can (why?)
The purpose of this was to find a
can that would have minimum cost
for materials (“tin”) used to make
the can.
To the right are two parts of the
entry for “Tin can” in Wikipedia.
We will investigate the first
sentence under “Standard sizes” in
this project.
Recall the problem was set up as:
=
Volume: given to be .
•
so we get the relation
ℎ
ℎ= ⁄
Surface Area = (area of top and bottom circle) + (rolled-out area of middle)
•
=2
+2
ℎ
ℎ
We then reduced the formula for to one in a single variable
=2
•
+
and found the derivative to be zero when
=
⁄2
/
=2
/
.
⁄
From the second-derivative test (
> 0) and the end-behavior (limits at → 0
and → ∞), we see that this critical point corresponds to an absolute minimum of .
Using the formula for ℎ above, we can relate ℎ to :
•
ℎ=
⁄
=
⁄
⁄2
⁄2
/
=2
= ⁄2
The radius of the top of the
can is half the diameter:
Can with ℎ =
So we conclude that the can that will hold volume that uses the minimum amount of
material will have a diameter that equals the height of the can.
But this can will look like a square if we look at its front profile. But no cans in a grocery
store look like a square from the front view. So what is the issue here?
PART I: MINIMUM MATERIALS Follow steps 1 through 3 below to solve the optimum can problem with Geogebra.
1. You can install Geogebra, a free open source (no advertisements, no malware, etc) computer algebra system, on
your computer or you can access it directly through a web browser (not as quick, but is easier to get started).
(a) go to the webpage www.geogebra.org/cms/en/download
(b) select the [Applet Start] button to run Geogebra without installing it on your computer; if you want to
install Geogebra, instead select [Webstart]
i. if you select [Applet Start] and get the message “the required plugin is not installed” or a message asking you to
“please go to www.java.com”, then you need to follow the 3 bullet items below to install “java”, which is a very
common program needed to run many applications:
•
•
•
go to www.java.com if you get this message, and click the [Free Java Download] button
on the next screen, make sure the operating system displayed is correct for your computer – if it is
correct, select [Agree and Start Free Download] button;
if not correct, click the link See all Java downloads and install the correct one
you will see a screen about a “FREE Browser Add-on”;
UNcheck the box below to the left of “Install the Ask Toolbar….” - this is junkware
We will step through the problem again here, with a slight twist to make it easier to compare the radius and the height
of the can. We will look at the ratio of the radius to the height so we don’t have to relate ℎ to .
For a cylindrical can of radius and height ℎ, we’ll consider the ratio
= ⁄ℎ .
2. Algebraically find
(a) Substitute
(b) Substitute
, the scaled version of
= ℎ into
to show
= ℎ into the formula for
= ℎ.
Then
′
, and find
=2 ℎ
and
+
′′
:
and solve for ℎ to show ℎ = "
&
#$ %
Also find as a function of .
(c) Use your result in (b) to find ℎ in terms of
(d) Write the equations for ′
and ′′
and , and substitute that into
=2
⁄
⁄
'
⁄
. Remember that
+
⁄
(
⁄
.
in (a) to show
is a constant.
3. Plot
. First, near the upper-left, click “Options” , then select “Rounding” underneath, and click “4 Decimal
Places” to the right. In the “Input:” bar at the bottom, type the following and hit Enter after each line/(letter).
Do not type comments in braces {} below!
(a) {below, we will draw grid lines in the graph area}
i. {point anywhere in the graph window and right-click. At the bottom, select “Graphics…”}
ii. {select the “Grid” tab and click on the check the box to the left of “Show Grid”}
(b) {type the lines below which we’ll use to adjust the view area of the graph}
i.
ii.
iii.
iv.
v.
x1 = −0.5
y1 = x1
x2 = 2
y2 = 9
Zoomin[x1, y1, x2, y2]
(c) V = 0.0425
{This will set the volume variable V to 0.0425 for now.}
(d) S x = 2 ∗ pi^ 1⁄3 ∗ V^ 2⁄3 ∗ x^ 2⁄3 + x^ −1⁄3 {Note Geogebra uses “x” for domain. After “(“ Geogebra automatically adds the closing parenthesis “)” –
this will plot the function S x , where x represents . }
i. {point to any part of the graph of S x , and right-click, then select “Object Properties…” under}
ii. {select the “Color” tab at the top, and choose one of the shades of red}
(e) S′ x = Derivative[S, x]
{Note the ‘prime’ after S (this is the single-quote character). This will plot the derivative, follow steps (i) and (ii) in part
(c) above, but choose a shade of blue. If you don’t have a color printer, under the “Style” tab at the top, choose the
second long-dash line style.}
(f) S′′ x = Derivative[S′, x]
{Note the ‘double prime’ after S (this is two single-quote characters). This will plot the derivative, follow steps (i) and
(ii) in part (c) above, but choose a shade of purple. If you don’t have a color printer, under the “Style” tab at the top,
choose the third short-dash line style. }
(g) {Label the graphs}
i. { Right-click on the graph of S x and choose [Object Properties…] below. Click the [Basic] tab at the top, and
click on the box to the left of “Show Label:” so it has a check mark. Click the box [Name v] and select “Name &
Value”, then click [Close] at the bottom. This will label the graph with S x .}
ii. Do the same for the graphs of S′ x and S′′ x . You can click and drag the labels to move them.
(h) P = min[S, 0, 5]
i. {This will plot the point P at the minimum value of the function S x . Right-click on the point P and select
“Object Properties…” at the bottom of the menu. Then select the “Basic” tab and click the box [Name v] below
and choose “Value”. Then click [Close] below – this will display the coordinates of the critical point. Drag the
label to move it so it is not obstructed by the graph.}
(i) {Now we’ll find the zero of the derivative S′ x to verify the point P is a critical point. Type the following:}
i. xmin = x P
ii. Q = Point[Gxmin, S H xmin I]
iii. {Follow step (i) under (h) above to label the point Q }
(j) {Print your graph: (do not type what’s below, only follow the steps) }
i. Click in the graphics area to activate it so it prints. Then in the upper-left of the Geogebra window,
click “File” and select “Print Preview” underneath. Add your name (Author), a title, and click “Print”.
(k) {Explain what the coordinates of the point J represent from (d) (do not type what’s below) }
i. Use the formula and graph for S′ to verify P is a critical point of S.
ii. Use the graph and formula for S′′ to verify P is a local minimum of S (2nd derivative test).
iii. Use the graph and formula for S′ to justify that P is a global/absolute minimum of S. Hint: think about
the end-behavior of S x . If S′ x is positive (negative) on an interval, then S x is increasing
(decreasing). Can S x ever go below the point P?
iv. Explain, in clear, correct, and complete sentence(s) what the coordinates of P represent in the context
of the problem with the can, such as if you had to explain your results to an owner of a can-making
company who didn’t know much about math. Is this consistent with the Wikipedia article?
Calculus I (Sem1)
Dr M. Campbell
Project 2: Finding an Optimal Can
PART II: EFFECT OF RECTANGULAR STORAGE
Now we will look at the influence of the rectangular nature of storage. The cabinets in which we put cans are
rectangular, as are the boxes in which they are shipped, as well as the cargo trailers on the 18-wheel semi-trucks that
are used to ship canned goods to grocery stores. For simplicity, we’ll only investigate the effects of the shipping/storage
cost that result from having to ship cans to a grocery store or the cost of real estate for the storage space for the cans.
To this end, we need to look at the feasible range of can sizes, the dimensions of a common cargo trailer for shipping or
cupboard for storage, and the cost of a truck driver shipping the cans or the cost of real estate for storing cans.
II.A Formula for the Material Cost to Make a Can
Table 1: Dimensional Can Standards
The table to the right was (mostly) obtained from the Can
Manufacturers Institute
(http://www.cancentral.com/standard.cfm#foodcan).
We’ll take the thickness of a can to be about 0.2032mm or,
equivalently, about 0.008in or about 6.67 ⋅ 10 O ft ***.
diameter height
Name
of Can
We need the cost of a can, which we can get from the following:
6Z
8Z
Density:
7850 kg⁄m **
Short
or 490.1 lb⁄ft
8Z Tall
No. I
Cost: US $800 - $1,050 per Metric Ton *
(Picnic)
Bulk of sheets of tin plate No. 211
or $0.363 - $0.476 per pound
Cylinder
steel used to make cans
No. 300
*** www.alibaba.com/product-gs/656173961/metal_food_can.html
No. 300
** www.emirapackaging.com/english/ASTM-A624M-84.htm
Cylinder
* www.alibaba.com/product-gs/555207963/tinplate_for_two_piece_can.html
No. I
For simplicity, we’ll assume that the cost of
Tall
• making a can is proportional to the volume of tin to make the can, No. 303
No. 303
• tin plate steel is $0.40 per pound from * above.
1. Use the cost of tin plate steel above to compute the cost per Cylinder
No. 2
square inch of surface area to make a can:
Vacuum
• a sheet of tin plate steel will be 0.008 inches thick, so the volume No. 2
of tin to make a can is the can’s surface area
times 0.008
Jumbo
(think of lying flat the top, bottom, and rolling out the side flat)
No. 2
• Justify that the cost is related to the surface area of the can by:
Cylinder
$b.Ob Oeb. cd
No.
O
material cost: _`
=
⋅
⋅ 6.667 ⋅ 10 ft ⋅
cd
fgh
1.25
No. 2.5
which reduces to
kl
$b. bi
$e.bij⋅ b
No. 3
_`
=
=
Vacuum
fg%
mn%
No. 3
where = ⁄ℎ , and
is given as square feet for the first
Cylinder
formula and square inches for the second formula on the right.
No. 5
Tin Plate Steel
Then the cost of materials is given by a constant times
.
No. 10
capacity
volume
Q = R S/T T U
Total
Capacity
2
2
3
8
avoir oz. = /U
of
VWX
YZ X
Water
at 68°F
12.41 0.0072 6.08 0.303571
2
11
3
0
17.02 0.0098
7.93
0.447917
2
11
3
4
18.44 0.0107
8.68
0.413462
2
11
4
0
22.69 0.0131 10.94 0.335938
2
11
4 14
27.65 0.0160 13.56 0.275641
3
0
4
7
31.37 0.0182 15.22 0.338028
3
0
5
9
39.32 0.0228
19.4
0.269663
3
1
4 11
34.53 0.0200
16.7
0.326667
3
3
4
6
34.91 0.0202 16.88 0.364286
3
3
5
9
44.39 0.0257 21.86 0.286517
3
7
3
6
31.32 0.0181 14.71 0.509259
3
7
4
9
42.34 0.0245 20.55 0.376712
3
7
5 10
52.20 0.0302
25.8
0.305556
3
7
5 12
53.36 0.0309
26.4
0.298913
4
1
2
6
30.79 0.0178 13.81 0.855263
4
1
4 11
60.76 0.0352 29.79 0.433333
4
4
3
7
48.77 0.0282
23.9
0.618182
4
4
7
0
99.30 0.0575
51.7
0.303571
5
2
5 10 116.04 0.0672
59.1
0.455556
6
3
7
S
U
S
U
1/16
1/16
in
in
in
in
0
210.48 0.1218 109.43 0.441964
The concept of cost allows us to connect the issues of minimum surface area to the issue of storage volume usage in a
shipping truck. We can’t compare square inches of a can to cubic inches used in a shipping truck (different units).
Finding the cost of each allows us to connect the concepts by adding their costs (same units: dollars).
II.B Formula for Storage/Shipping Cost: Unused Volume in a Shipping Truck
“A simple rule to keep in mind is the larger the quantity you have shipped at once,
the less you will pay per container for freight.” (http://www.skolnik.com/faq.php)
ℎ
Now we need to look at the distribution cost of volume for in a truck for choosing a
can of a specific radius and height. First we note a few points:
Simple Example:
• there will always be a necessary amount of unused volume
because a can does not have a rectangular top and
Internal and External Dimensions of a Cargo Trailer that Ships Cans
bottom (can opener requirement)
width
General
20'
• when we say “no wasted volume”, we mean no
Specifications
(length)
unnecessary unused volume; e.g., 2, 3, and 4, cans
height
below have unnecessary unused volume, which can be
Dimensions:
ft
in
corrected by lowering the diameter to 4 ft. (10 cans)
• we can see below that the dimensions of the cargo trailer
Length 19 4 3/16
will result in some radii and heights being more efficient
Internal Width 7
8 1/2
• we will only consider practical dimensions for cans in the
length
Height 7 10 3/16
range shown in Table 1 above
www.containersales.com/containers-for-sale-us/container-information-dry-cargo.aspx
We’ll look at a discontinuous function that gives the number outbackstoragecontainers.com/shipping-container-massachusetts
of cans along the length as a function of the diameter of the
Storage Possibilities Assuming External Dimension (for simplicity)
can. Note this doesn’t depend on the height of the can.
2. We define the following
• ℒ = length of the cargo trailer
• p = width of the cargo trailer
• ℋ = height of the cargo trailer
If we only consider length and don’t restrict width (such as
with the flatbed trailer above with only a bottom), the
number of cans along the length, rℒ , for a very large can
(i.e., we could only fit one can on
the cargo trailer) of diameter is
Case with One Can
u−
1. rℒ
=1
for
ℒ
<
≤ℒ
2. rℒ
=2
for
ℒ
<
≤
Four cans (5 ft diam)
•
no
wasted
volume along length
When the most that can fit are two
•
wasted
volume
along width
cans (see cases on right), we have
u⁄2
u
ℒ
3. rℒ
=3
for
ℒ
O
<
≤
ℒ
Let u = ⁄ ℒ/'2 ⁄ (. Then we get a function
vw
(in terms of , u) for the maximum number of
rℒ
cans we can fit along the length of the cargo trailer
for a given (see functions on right). Verify below:
<
≤
ℒ
%
ℒ
h
<
%xy⁄h
z y⁄h
⁄
≤
2≤u
ℒ
%
−1⁄3
<3
The maximum number of cans that fit along the
width, vƒ
, can be found similarly (to right).
Let „ =
p
<
≤
⁄
p
Ten cans (4 ft diam)
• no wasted volume along length
• no wasted volume along width
With three cans efficiently stored,
The maximum number of cans that fit along the length can
be found in terms of by using the constant (capacity)
volume of the can to get ⁄2 = = 5 ⁄ 7 ⁄ .
ℒ
h
Largest possible cans (8 ft diam)
Three cans (20/3 ft diam)
• wasted volume along length
• no wasted volume along length
• no wasted volume along width • wasted volume along width
p/'2
p
<
⁄
( and verify the following:
y⁄ h
# y⁄h
⁄
≤
p
2≤
„
1⁄3
<3
Ten cans (3 ft diam)
• wasted volume along length
• wasted volume along width
=
1
~
|2
}⋮
|
{€
ℒ
%
ℒ
h
<
ℒ
•‚y
<
<
≤ ℒ
⋮
≤
ℒ
≤
ℒ
`
Max. # Cans along Length
rp
=
1
~
|2
} ⋮
|
{…
p
%
p
h
p
†‚y
<
<
<
≤ p
⋮
≤
p
≤
p
‡
Max. # Cans along Width
vw
Ten cans (3 ft diam, 3.5 ft height)
• wasted volume along len/width
• wasted volume along height
=
1
~
|2
} ⋮
|
{€
1≤u
2≤u
€≤u
⋮
Max. # Cans along Length; u =
vƒ
=
1
~
|2
} ⋮
|
{…
<2
⁄
1≤„
<3
⁄
<€+1
⁄
⁄
ℒ/'2
⁄
2≤„
⁄
⁄
<2
€≤„
⁄
< €+1
⋮
Max. # Cans along Width; „ =
⁄
(
<3
p/'2
⁄
(
The maximum number of cans that fit along the height
1ℋ% < ℎ ≤ ℋ
~
can be found similarly (see to right). To get instead
ℋ
ℋ
|
of ℎ, we need to use the constant (capacity) volume of rℋ ℎ = 2 h < ℎ ≤ } ⋮⋮
] ⁄ . With ˆ = ⁄ ℋ/ ⁄ ,
the can ℎ = [ ⁄
ℋ
|
ℋ
verify the following:
{ Š‹‚y < ℎ ≤ Œ
ℋ
<ℎ≤ ℋ
ℋ >
# y⁄h
y⁄h
⁄
≥ ℋ
2 ≤ ˆ
⁄
<3
v•
=
11 ≤ ˆ
~
| 22 ≤ ˆ
< 2
⁄
< 3
⁄
}⋮⋮
|
{ ŠŠ ≤ ˆ ⁄ < Š + 1
Max. # Cans along Height Max. # Cans along Height ˆ =
⁄
ℋ/
⁄
Now we will look at the total shipping cost per volume of canned goods distributed by the cargo trailer. As an example, suppose you
need to deliver 300 cans of volume 2 in and you have to pay $150 (independently of the number of cans) to a truck driver to make
a single delivery trip to a grocery store. Suppose the can size is efficient and you can fit all 300 cans in the truck. Then the shipping
cost per volume of goods delivered is $150⁄ 300 ⋅ 2 = $0.25 per in . If, because of an odd can size with the same volume 2 in ,
you can only fit half the number of cans in the truck, then only half the volume of goods are delivered for a shipping cost (per unit
volume) of $150⁄ 150 ⋅ 2 = $0.50 per in . If yet another odd can size only allows you to fit one-third the original amount in the
truck, the shipping cost (per unit volume) would be $150⁄ 100 ⋅ 2 = $0.75 per in .
To get the actual cost, the owner (a friend of mine) of Roadmaster Transport, a
52’ = 624”
trucking service and transportation company in Santa Fe Springs CA, was contacted to
100”
obtain the costs associated with truck delivery of canned goods. He mentioned that
the most common semi-truck for delivery is a 53 foot-long “dry van” (picture at the top
110”
of the previous page). Furthermore, he stated the cost to ship any amount of dry
goods is a per-delivery flat fee of $150. Lucky for us, this keeps the problem relatively
simple, and we will consider the 53-foot cargo trailer since it is the most common and
Interior Dimensions of 53 ft Dry Van
would have the biggest influence on can design to minimize unused volume.
length = 624 in, width = 100 in, height = 110 in
150
150⁄
=
⋅v
v
3. Show clearly and justify that the formula for the shipping cost per volume of canned goods (per trip) is :
_Ž =
where is the capacity-volume of a single can and v is the maximum number of cans that will fit in the cargo trailer. Note that
the second expression 150⁄ ⁄v is a per can quantity, and we will combine this cost with the material cost per can below.
The cost of shipping is a function of :
_Ž
=
150⁄
,v
v
= vw
vƒ
v•
4. Find the optimal can dimensions that minimize the combined material and shipping costs for the No. 2 can and the $150
shipping cost for the 53-ft Dry Van. Note the blue graph, Cm, is the graph of the material cost function from the previous
project. Enter the following in Geogebra:
i. Can Dimensions
diam= 3+7/16
height= 4+9/16
ii. Truck Dimensions iii. Can Info
Cost Tin…=$ 0.0009076
length= 624
Cost Distribution…=$ 150
width= 100
iv. Display Window
x1= 0.48
x2= 0.52
y1= 0.06103
height= 110
Print your graph. Add your name and a title.
y2= 0.06108
5. Try using the cost of storing cans for a month in a cupboard in Tokyo, Japan. The average rent (Sep 2012) is 2,995 Yen per
square meter. We can convert that to dollars per square foot with dimensional analysis. Since we can stack cupboards
vertically, we need to look at the rent in dollars per volume. We will assume the ceilings are an average of 8 ft high, so the cost
of rent per volume can be computed as follows (c.f., “Apartment Rents Reach 7 Year Low in Tokyo”, Japan Property Central, 5
Sep 2012):
Tokyo, Japan (Sep 2012) – Average Condo Rent ($ per sq ft) : 2,995
Average Condo Rent ($ per cu ft):
$ . ‘
fg%
⋅
¥
•%
‘fg
⋅
•%
⋅
$
10.ij e fg% ‘ . ’¥
= 0.42235
$
fgh
= 3.3788
$
fg%
Compare that to the cost for retail store rent in Manhattan (New York) on lower Fifth Avenue between 42nd and 49th streets,
which runs an average of $900 per square foot (in Spring 2012, c.f., “The Heart of Fifth Avenue Shopping is Heading to the
South”, N.Y. Times, 4 Sep. 2012), so we get
Manhattan, New York (Spring 2012) – Average Retail Rent ($ per cu ft) :
$ebb
⋅
fg%
‘fg
= 112.50
$
fgh
We can compare this to the $150 cost of storage in the 53-ft cargo trailer, since the cost per month to store these cans in a
cupboard is a flat per-month fee based on the rent per cubic foot. For example, a cupboard that measures 4 ft in length
(across), 1 foot in width (distance to back of cupboard), and 1 foot high has a volume of 4 ft . So the amount of money the
tenant in a retail shop on Lower Fifth Avenue would be 4ft ⋅ 112.50 $⁄ft = $450 per month! So we can consider that to be
equivalent to the Cost of Distribution ($ per Delivery).
(a) Enter the optimal can size from the 53-ft Dry Van problem for the can dimensions and see what effect the cost of rent
for cupboard space in Tokyo would have by entering the dimensions for the cupboard under “Truck Dimensions” and
the per month cost of rent per cubic foot under “Cost Distribution ($ per delivery)”. You can move point coordinates
around so they are not covering the graphs.
i. Can Dimensions
diam= 3.7821
height= 3.769
ii. Truck Dimensions iii. Can Info
iv. Display Window
Cost
Tin…=$
0.0009076
x1= 0.26
x2= 1.0
length= 36
Cost
Distribution…=$
1.6894
width= 12
y1= 0.06
y2= 0.064
height= 12
Print your graph. Add your name and a title.
(b) Enter the optimal can size from the 53-ft Dry Van problem for the can dimensions and see what effect the cost of rent
for cupboard space in Retail Manhattan would have by entering the dimensions for the cupboard under “Truck
Dimensions” and the per month cost of rent per cubic foot under “Cost Distribution ($ per delivery)”. Note that you
only need to change a few fields. You can move point coordinates around so they are not covering the graphs.
i. Can Dimensions
diam= 3.7821
height= 3.769
ii. Truck Dimensions iii. Can Info
iv. Display Window
Cost
Tin…=$
0.0009076
x1= – 0.125
x2= 1.0
length= 36
Cost
Distribution…=$
112.50
width= 12
y1= – 0.1
y2= 0.125
height= 12
Print your graph. Add your name and a title.
point on the graph by zooming in. Note that a calculation with a calculator shows = 0.2504 which does in fact
match Geogebra’s calculated cost to one in one-millionth of a dollar. So we see that rectangular storage, when small
enough and expensive enough, does affect the size of a can.
(c) Notice that it does not look like there are any points on the graph at the point PMin. Double check that there is a
i. Display Window
x1= 0.02495
x2= 0.251
y1= – 0.02
y2= 0.16
Print your graph. Add your name and a title.
6. Is the can size affected by distribution cost of a 53-ft Dry Van? Test a few other can sizes to see the effects of storage by using
the app to calculate the optimal diameter and height from the
values obtained. Justify your answer with your results.
7. Does the smaller 20-ft storage trailer at the top of page 2 have any effect on dimensions? Justify your answer with some
results.
8. Is the can size affected by real estate costs for storage? Justify and explain your results with data for both the Tokyo and New
York case for a few can sizes.