Millions of cans and ice cream cone packaging are produced and consumed daily. Consuming these products myself, I’m shocked at the amount of waste just one person can produce. Reducing waste is often talked about as a strategy to reduce pollution and the effects of climate change and one seemingly obvious way to do this would be to optimise packaging, reducing the amount of material needed to contain a certain volume of product. Thus the aim of this investigation is to look at two common shapes, cylinders and cones, and to find the optimal ratios of radii and height which minimizes the surface area, thus reducing packaging to look at whether current manufacturing has observed these ratios to reduce packaging.
Consider a commonly shaped object like a can. Although usually not a perfect cylinder, for this investigation, they will be considered perfect cylinders
The formula for the volume of a cylinder is as follows -
\(V- πr^2h\)
This can be rearranged to make h the subject (Although both r and h would work, it was decided that making h the subject would simplify the process since making r the subject would involve a square root)
\(h= \frac{V}{πr^2}\)
The surface area of a cylinder can be calculated by summing the areas of the rectangle forming the connecting piece between the two circles, and the area of the two circles.
SA = 2πr^{2 }+ 2πr h (Equation 2)
Equation 1 can now be substituted into Equation 2 to get SA in terms of r and V
\(SA = 2πr^2 +2πr\frac{V}{πr^2}\) (Equation 3)
Now, \(\frac{d(SA)}{d(r) }\)can be computed while keeping V constant to find the turning point of the function. V must be kept constant as this is part of the research question. The power rule was used at this stage, the rule is as follows -
f ' (x)^{n} = nx^{ n-1}
Therefore,
\(\frac{d(SA)}{d(r)} = (2)2πr^{(2-1)}+(-1)2Vr^{(-1-1)}\)
\(\frac{d(SA)}{d(r)}=(2)2π^{(2-1)}+(-1)2Vr^{(-1-1)}\)
\(\frac{d(SA)}{d(r)}= 4πr-2Vr^{(-2)}\)
In order to find the minimum point, Equation 4 was equated to 0, giving the turning point of the function.
0 = 4πr - 2Vr ^{-2}
2Vr -2 = 4πr
V = 2πr^{3}
Since V has maintained constant, Equation 1 can be substituted for volume
πr^{2 }h = V = 2πr3
h = 2r (Equation 5)
It has been established that a turning point occurs when h = 2r, however, this value could be a maximum or a minimum. The second derivative of Equation 3 can be used to show that this is a minimum value. The second derivative is the rate of change of the gradient, if the second derivative is positive, the graph will be concave up, meaning the point will be a minimum.
\(\frac{d^2(SA)}{d(r)^2}=\frac{d(SA)}{d(r)}4πr-2V^{-2}\)
\(\frac{d^2(SA)}{d(r)^2}=4π+4Vr^{-3}\)
It can be said that V ϵ N+ and r ϵ N+ as these are real measurements that can be taken in the physical world, this ensures that the equation above must be positive, thus the point is a minimum, demonstrating that the ratio in Equation 5 will optimise for the minimum surface area.
The surface area of this can is given by Equation 2, where h is 12.2 cm and r is 6.62 cm.
SA = 2π (3.31)^{2 }+ 2π (3.31) (12.2)
SA = 323 cm^{2}
The volume of a can of this nature is 355 cm3. Based on this fixed volume, the optimal ratio between h and r can be used to determine the minimum surface area. Equation 5 can be substituted into Equation 1
355 = rπ2(2r)
r_{opt }= 3.84 cm
This value can be substituted into equation 3
\(SA=2π(3.84)^2+\frac{2(355)}{(3.84)}\)
SA _{opt} = 278 cm2
The wastage can be given by
SA_{Actual }- SA_{Opt}
323 - 278 = 45 cm2
Thus for each can of these dimensions, 45 cm2 of material is wasted, or 16.2%. Each aluminium can weighs about 14 grams, assuming a percentage wastage of 16.2% in the surface area carries over to volume due to its thin nature, 2.24 grams of aluminium is wasted per can. According to TheWorldCounts, 129 billion cans have been produced this year, giving a total wastage of almost 290,000 tons of aluminium.
The same processes can be repeated to determine the optimal ratio of height versus radius.
\(V_{cone}=\frac{πr^2h}{3}\)
\(h=\frac{3V}{πr^2}\) (Equestion6)
The equation for the surface area of a cone can be derived using a net (Figure x) As seen in the net below, the sector of the circle folds to fit around the base circle. Thus it can be said that the circumference of the sector is equal to the circumference of the circle.
\(Circumference_{Sector}=2r \)
It can also be said that the radius of the sector of the circle is equal to the diagonal length from the circumference of the base of the cone to the vertex. Thus, using Pythagoras’ theorem,
R^{2} = h^{2} + r^{2}
^{\(R=\sqrt{h2+r2}\)}
In order to determine the area of the sector, the proportion of the circle contained within the sector must be known. This can be found by multiplying the total area of the circle and the ratio of the circumference of the sector to the circumference of the whole circle.
\(A_{sec}=\frac{2π r}{2π \sqrt{h^2+r^2}}×π \sqrt{h^2+r^2}\)
\(A_{sec}=π r\sqrt{h^2+r^2}\)
The total surface area can be found by adding the area of the sector to the area of the small circle radius r.
SA_{Cone} = A_{Sec} + A_{Circle}
_{\(SA_{Cone}=rπ \sqrt{h2+r2}+π r^2\)} (Equation 7)
Equation 6 can then be substituted into Equation 7
\(SA_{cone}=rπ\sqrt{ \frac{(3Vr2)}{2}+r2}+rπ 2 \)
\(SA_{cone}=rπ \sqrt{\frac{9V^2}{π ^2r^4}+r2}+rπ 2\) (Equation 8)
This equation was then differentiated with respect to r, substitution was used to simplify this process. Consider
\(u=\frac{9V^2r^{-4}}{π 2}+r^2 \)
Therefore,
\(u'=\frac{(-4)9V^2r^{(-4-1)}}{π 2}+2r^{(2-1)} \)
\(u'=-\frac{36V^2r^{-5}}{π 2}+2r \)
The product rule and the chain rule were used to differentiate the equation above. The product rule is shown below -
\(\frac{d(uv)}{d(x)}=\frac{ud(v)}{d(x)}+v\frac{d(u)}{d(x)} \)
The chain rule is shown below -
\(\frac{d(y)}{d(x)}=\frac{d(u)}{d(x)}×\frac{d(y)}{d(u)}\)
Applying these rules,
\(\frac{d(SA)}{d(r)} = \pi r \left(u^{\frac{1}{2}}\right)' + \pi u^{\frac{1}{2}} + 2\pi r \)
The chain rule can be applied to find \((u^{1/2})’\)
\(\left(u^{\frac{1}{2}}\right)' = \frac{1}{2}u^{\frac{-1}{2}}×u'\)
\(\left(u^{\frac{1}{2}}\right)' = \frac{u'}{2u^{\frac{-1}{2}}}\)
This can be substituted into the equation above
\(∴ \frac{d(SA)}{d(r)}=\frac{rπ u'}{2u^\frac{1}{2}}+π u^\frac{1}{2}+2π r \)
This can be equated to 0 to find the turning point
\(0=\frac{π ru'}{2u^\frac{1}{2}}+π u^\frac{1}{2}+2π r\)
Since is a common factor, both sides can be divided by to simplify the equation above
\(0=\frac{ru'}{2u^\frac{1}{2}}+u^\frac{1}{2}+2r \)
To distribute \(u^\frac{1}{2}\), both sides of the equation were multiplied by \(u^\frac{1}{2}\)
\(0=\frac{ru'}{2}+u^\frac{1}{2}(u^\frac{1}{2})+2r(u^\frac{1}{2})\)
\(0=\frac{ru'}{2}+u+2ru\frac{1}{2} \)
While both sides could have been multiplied by \(2u^\frac{1}{2}\), the denominator, 2, cancels nicely in the following steps, making the equation more simple. u’ and u can now be substituted into the equation above.
\(0=\frac{r(\frac{-36V^2}{π ^2r^5}+2r)}{2}+(\frac{9V^2}{π ^2r^4}+r^2)+2r(\frac{9V^2}{π ^2r^4}+r^2)^\frac{1}{2}\)
\(-\frac{r\frac{-36V^2}{π ^2r^5}+2r^4}{2}+\frac{9V^2}{π ^2r^4}+r^2+2r(\frac{9V^2}{π ^2r^4}+r^2)^\frac{1}{2}\)
\(\frac{18V^2}{π ^2r^4}-r^2=\frac{9V^2}{π ^2r^4}+r^2+2(\frac{9V^2}{π ^2r^4}+r^2)^\frac{1}{2}\)
\(\frac{9V^2}{π ^2r^4}-2r^2=2r(\frac{9V^2}{π ^2r^4}+r^2)^\frac{1}{2}\)
\((\frac{9V^2}{π ^2r^4}-2r^2)^2=(2r(\frac{9V^2}{π ^2r^4}+r^2)^\frac{1}{2})^2\)
\(\frac{18V^4}{π ^4r^8}-r^2=\frac{36V^2}{π ^2r^2}+4r^4=4r^2(\frac{9V^2}{π ^2r^4}+r^2)\)
\(\frac{18V^4}{π ^4r^8}-\frac{36V^2}{π ^2r^2}+4r^4=\frac{36V^2}{π ^2r^2}+4r^4\)
\(\frac{18V^4}{π ^4r^8}=\frac{72V^2}{π ^2r^2}\)
9V^{2 }= 8π ^{2}r^{6}
^{Equation 6 can be substituted into the equation above.}
^{\(9(\frac{(rπ^ 2h}{3})^2=8π ^2r^6\)}
h^{2 }= 8r^{2}
^{\(h=2\sqrt2r \)}
Since the method was identical to the previous shape and since that was proven to be a minimum, it was initially assumed that this ratio would also yield a minimum surface area. This is addressed later on in the exploration.
A classic Cornetto ice cream cone has the following dimensions -
r = 3.3cm
h = 17.0cm
The surface area of the cone can be calculated using Equation 7.
\(SA_{cone}=π(3.3)(17.0)+π(3.3)\sqrt{17.0^2+3.3^2}\)
SA_{cone }= 356 cm^{2}
The volume of the cone can be calculated using Equation 6
\(V_{cone}=\frac{π(3.3)^2(17.0)}{3} \)
Vcone = 194cm^{3}
^{For this volume, the optimal surface area can be found by substituting the optimal ratio into Equation 6.}
^{\(194=\frac{πr2(2\sqrt2r)}{3}\)}
r_{opt} = 4.03cm
∴ h_{opt} = 11.4cm
\(∴ S_{Aopt}=π(4.03)^2+π(4.03)^\sqrt{4.03^2+11.4^2}\)
S_{A}_{opt} = 204.1cm^{2}
Thus the wastage is
356 - 204 = 152cm^{2}
Giving a percentage wastage of
\(\frac{152}{356}×100 = 42.7%\)
This is an extremely large percentage, creating hundreds of tonnes of waste.
The same method to prove the minimum was attempted. This involved taking the second derivative of Equation 8 with respect to r and proving it was positive. This resulted in an extremely long inequality. In order to avoid this, I attempted to justify the minimum by trialling other ratios with a fixed volume. For example, using the ratio h = 2r
\(V_{cone}\frac{=πr2h}{3}\)
\(194=πr^\frac{2(2r)}{3} \)
r = 4.52cm
\(SA=πr\sqrt{(2r)^2+r^2}+πr^2\)
\(SA=π4.52\sqrt{(24.52)^2+4.52^2}+π4.52^2\)
SA = 207.7cm^{2}
Ratio | Surface area |
h = 2r | 207.7 |
\(h = 2\sqrt{2r }\) | 204.1 |
h = 3r | 204.3 |
While this shows that the ratio \(h=2\sqrt{2r}\) is the minimum among the three trialled, it does not show it is the overall minimum of the function. In order to determine this, a new method was taken -
Consider the ratio between h and r a variable, x.
\(\frac{h}{r}=x\)
h = xr (Equation 9)
Consider a fixed volume
\(V=194=\frac{πr^2h}{3} \)
Equation 9 can be substituted into the equation above
\(194=\frac{πr^2xr}{3} \)
\(\frac{1943×}{πx}=r^3\)
\(r=\sqrt[3]{\frac{582}{πx}}\)
This can be substituted into Equation 8
\(SA_{\text{cone}} = \pi \sqrt[3]{\frac{582}{\pi x}} \sqrt{\frac{9 v^2}{\pi^2 \left( \sqrt[3]{\frac{582}{\pi x}} \right)^4} + \left( \sqrt[3]{\frac{582}{\pi x}} \right)^2 + \pi \left( \sqrt[3]{\frac{582}{\pi x}} \right)^2} \)
This gives surface area in terms of x with everything else being known constants. This can be plotted on a graph as shown below
This minimum point describes the ratio (x value) which correlates to the minimum, positive, surface area (y value).
\( 2\sqrt2\simeq 2.828 \)
Thus showing that the ratio is both correct and a minimum.
The same method can be applied to the cylinder. Equation 9 was substituted into Equation 1 -
V = πr^{2}xr
For the fixed volume 355cm3 (volume of a generic soda can), r can be found in terms of x -
\(r=\sqrt[3]{\frac{3355}{πx}}\)
This can be substituted into Equation 3 -
\(SA=π2(\sqrt[3]{\frac{355}{πx}})^2+\frac{2(355)}{\sqrt[3]{\frac{355}{πx}}}\)
This can be plotted on a graph of SA versus x -
The local minimum indicates the optimal ratio (x value) and the minimum surface are associated with this ratio assuming a fixed volume of 355cm3. This proves the initial algebraically calculated optimal ratio to be true.
This investigation was successful. The optimal ratio between the height and radius to reduce surface area while maintaining the volume of cones and cylinders was found using two methods: algebraically and graphically. The initial method of optimization worked well for the cylinder as the initial formulas were more simple, whereas the method was doable but much more complicated for a cone due to more complex initial formulas. A new method was
investigated and used for the cone, yielding the same results showing both are feasible methods.
While this investigation used mathematics to determine percent wastage for cans and cornetto ice cream cones, it does not take into account human factors like ergonomics and appeal. The optimal ratio would yield two shapes with proportions which look odd and may not be ergonomic as shown below -
This is likely the explanation as to why manufacturers decide to use proportions outside the optimal, as the benefit of an appealing shape outweighs the cost of extra material.
“Aluminium Cans Facts.” The World Counts, 2022,
https://www.theworldcounts.com/challenges/consumption/foods-and-beverages/alumi
nium-cans-facts. Accessed 5 December 2022.