Predicting the amount of M&Ms that can fit Into a plastic container

Personally, as a student, I would always pack my candies or snacks when I am going out for a long day. I normally pack these candies into a plastic container from a previous batch of bubblegum purchased, to me this volume of the container holds the perfect amount of candies that would last for a week depending on the candy.

From all the variety of candies, M&Ms remain my favourite confectionery. However, when I purchase the 180g M&M packet to fill in the plastic container; there always would be quite an amount of M&Ms left that could not be filled into the container. If I do purchase smaller bags, it would result in the need to purchase more bags of M&Ms which would eventually raise the overall cost. This problem is commonly found in business, where optimizing the quantity of the product is crucial to reduce cost and increasing the efficiency of the product line. The main target of optimization in business is to minimize the cost to increase revenue (Maverick, 2023) I have decided to investigate this problem to predict the maximum amount of M&Ms that would fit into the plastic container I use to package my M&Ms.

In this investigation, I aim to use geometry to find the volume of the plastic container and integral calculus to calculate the volume of M&M to find the maximum amount of M&M that could fit in the container. I would count the actual number of M&Ms that could fit into the container to verify my calculations.

The M&M has an ellipsoid shape, specifically an oblate spheroid that resembles a squashed sphere. It is a surface of revolution where the shape is found by rotating the elliptical shape on its minor axis (short axis) (Weisstein, Eric W. "Oblate Spheroid.", 2023) To find the volume of the M&M, I would be using the volume of revolution rotating on the x-axis shown in the investigation.

 

\([volume \ of\ revolution = \pi \displaystyle\int^b_a y^2dx]\)

 

The full volume of the elliptical shape can be visualised as a solid of revolution is generated by rotating the ellipse shape around a horizontal axis. This is done by determining the axis of rotation and the limits of the integration. For this investigation, I would be assuming the M&M is a perfect elliptical shape to calculate the approximate volume of an M&M.

For this investigation, I would assume the M&M is -

• Perfect circle from the top view
• Perfect ellipse from the side view
• Overall a perfect elliptical shape

This allows me to calculate the approximate volume of the M&M.

• All measurements in this study are using centimetres (cm)
• Figures that are not otherwise stated are all created by the student

The following symbols denote specific variables used in this study -

• All calculations will be rounded to 5 decimal places except for values regarding the amount of M&Ms present.

I will calculate the internal volume of the container seen in Figure 2 by measuring the dimensions, which is done with a ruler.

First, I measured the diameter of the container on both sides to ensure it was equal to a square. The diameter of the container is 5.0cm± 0. 05. To measure the top's surface area with rounded corners, I will combine all 4 rounded corners to form a circle (Circle A). The rounded corners are divided into 4 showing a quarter of Circle A . The red dotted lines shown in Figure 4 are plotted to scale the quarter of the circle with a radius of 1cm, I uploaded the picture in Desmos

Second, I will find the dimensions of the corner square. The corner square shows a quarter of a circle of rounded corners. The circle's radius is 1.0cm, and the empty space between the quarter of the circle and square B will be shown as R. R will be subtracted from the surface area of the plastic container to find the actual surface area with rounded corners.

The square shown in Figure 5 is known as square B. In order to find the area of R, I will find the area of the square and the area of the quarter of the circle. Then subtract the area of the square from the area of the quarter of the circle.

 

Area of the square B -

area of square B = a²

 

area of square B = (1. 0cm)²

 

area of square B = 1. 000000cm²

 

Area of a quarter of circle A -

 

area of circle A = πr²

 

➔ Substitute the radius of square B to find the area of the circle A

 

area of Circle A = π(1. 0cm)²

 

area of circle A = 3. 14159cm²

 

➔ Divide Circle A by 4 to find the area of a quarter of circle A

 

area of a quarter of circle A = \(\frac{3.14159}{4}\)cm²

 

area of a quarter of circle A ≈ 0. 78540cm²

 

^{Area of R -}

 

area of R = area of square B − area of a quarter of a circle

 

area of R = 1. 00000cm² − 0. 78540cm²

 

area of R ≈ 0. 21460cm²

 

^{Area of Square A -}

 

area of a square = width × length

 

➔ The actual area of square A can be derived by subtracting the area of R from all 4 corners as R is not a part of the area of square A.

 

area of square A = w × length − 4(area of R)

 

area of square A = 5. 0cm × 5. 0cm − 4(0. 21460cm²)

 

area of square A = 25cm² − 0. 85840cm²

 

area of square A ≈ 24. 14160cm²

To find the internal volume of the plastic container, I measured the internal height of the container. I used a piece of paper to measure the internal height of the container and made a pencil to mark the height of the container. Then I used a ruler to measure the height. The height of the plastic container is 8.2cm ± 0. 05

Volume of rounded cuboid = surface area of base × height

 

➔ The equation below can be derived, utilising the equation above and the surface area of square A previously found

 

volume of V_c = area of square A × h

 

volume of V_c = 24. 14160cm² × 8. 2000cm

 

volume of V_c ≈ 197. 96112cm³

I will calculate the volume of an M&M (V_m) seen in Figure 8 by measuring the dimensions of m the M&M using a calliper.

Next, I measured the diameter of the M&M of its width and height. The width of the M&M is 13.8mm and the height of the M&M is 6.75mm, converted to cm, the width is 1. 38cm and the height is 0. 675cm. In order to find an equation representative of the shape of an M&M from a side view centred from (0,0). I will substitute the width and length measured respectively into x and y of the standard equation of an ellipse to result in a representative equation of an M&M.

 

➔ S_e = standard equation of an ellipse

 

\([S_e:\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1]\)

 

\(a=\frac{1}{2}\ width:b=\frac{1}{2}height\)

 

\(a=\frac{1}{2}1.38cm\)

 

a = 0. 69000cm

 

\(b=\frac{1}{2}0.675cm\)

 

b = 0. 33750cm

 

First, to show that the equation attained above models the shape of the M&M, I uploaded a picture of the M&M with a comparison of the standard equation of an ellipse derived above on Desmos .

Figure 12 shows that the side view of the M&M fits the standard equation derived for an ellipse.

 

➔ The term ellipse arc from this point onwards is used to represent the outline for a quarter of the M&M cross-section as shown in Figure 12

Therefore, I will proceed to calculate the ellipse arc of the M&M from a side view with standard equation of an ellipse

 

\([S_e=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1]\)

 

\([S_e=\frac{x^2}{0.69000^2}+\frac{y^2}{0.33750^2}=1]\)

 

\(\frac{x^2}{0.4761}+\frac{y^2}{0.11390625}=1\)

 

\(\frac{x^2}{0.4761}=1-\frac{y^2}{0.11390625}\)

 

\(x^2=0.4761(1-\frac{y^2}{0.11390625})\)

 

\(x^2=0.4761(1-\frac{1}{0.11390625}y^2)\)

 

\(x^2=0.4761-\frac{0.4761}{0.11390625}y^2\)

 

x² = 0. 4761 − 4. 17975y²

 

➔ The x is left as x²for calculation convenience in the next step

 

The equation derived above can be substituted into the volume of revolution equation. I will be applying the volume of revolution by rotating the equation around the x-axis to first find the volume of half the M&M.

 

\([\text{volume of revolution in y axis} = \pi\displaystyle\int^b_a \ y^2dy]\)

 

\(half \ of\ V_m =\pi\ \displaystyle\int^{0.3375}_0[0.4761-4.17975y^2]dy\)

 

➔ Finding the integral

 

\(half \ of\ V_m =\pi\ [0.4761y-4.17545× \frac{y^3}{3}]^{0.3375}_0\)

 

➔ Applying the limits

 

\(half \ of\ V_m =\pi\ [0.4761(0.3375)-4.17545× \frac{(0.3375)^3}{3}]\)

 

half of V_m = π[0. 16068375 − 0. 0535061083]

 

half of V_m = π(0. 10712)cm³

 

half of V_m = 0. 33653cm³

 

➔ Multiplying the value above by 2 to find the full volume of V_m

 

V_m = 0. 33653cm³ × 2

 

V_m = 0. 67306cm³

 

Therefore, the volume of the M&M is 0.67306cm³.

Calculating the Maximum Amount of M&Ms that Can Fit into the Container

 

The easiest way to predict the amount of M&Ms that can be found in my container, would be to divide the volume of the container by the volume of the M&M, this gives the maximum amount of M&Ms that could fit into my container.

 

Maximum Amount of M&Ms that can Fit = V_c ÷ V_m

 

Maximum Amount of M&Ms that can Fit =197. 96440cm³ ÷ 0. 67306cm³

 

Maximum Amount of M&Ms that can Fit ≈ 294. 12593

 

➔ As the answer has decimal places, the final value is round down to the nearest ones assuming the M&Ms are exist as a whole object with no damage

 

Maximum Amount of M&Ms that can Fit ≈ 294

 

The question arises, when the M&Ms are packed, they will have gaps between them. Following the calculation above, the M&Ms would be squashed together, which is not ideal for packaging products. This brings me to my next part, calculating the voidage between M&Ms to more accurately predict the amount that would be present.

 

➔ All values regarding the amount of M&Ms would be rounded down to the nearest ones assuming the M&Ms exist as a whole object with no damage.

The number of M&Ms assuming all M&Ms are identical in shape and size that are packaged can be modelled by the density of the package. The equation to calculate the amount of M&Ms able to fit into the container using packing density is shown below -

 

V_c × packing density of M&M = amount of M&M per cm³

 

amount of M&M per cm³ ÷ V_m = amount of M&M in plastic container

 

According to (Eric Weisstein. "Ellipsoid Packing.", 2023 ) the package density of identical ellipsoids have been calculated by Bezdek and Kuperberg, J.wills and Aleksandar Donev. Each value derived has been calculated under different circumstances; therefore the student would use the average of the calculated amount of M&Ms able to fit in the plastic container using each value given.

 

➔ All value of the amount of M&M calculated would be rounded down to the nearest ones, assuming the M&Ms are not damaged and exist as a whole.

• Bezdek and Kuperberg (1991)

 

 \(\text{Bezdek and Kuperberg (1991)}\frac{(24\sqrt{2}-6\sqrt{3}-2\pi)\pi}{72}=0.7533554605\)

 

197. 96440 × 0. 75336 = 149. 13846

 

149. 13846 ÷ 0. 67306 = 221. 58271

 

≈ 221

 

• Donev et al. (2004)

197. 96440 × 0. 68000 = 134. 61580

 

134. 61580 ÷ 0. 67306 = 200. 00563

 

≈ 200

 

• J.Wills (Bezdek and Kuperberg (1991))

197. 96440 × 0. 75850 = 150. 15600

 

267. 5135135 ÷ 0. 67306 = 223. 09452

 

≈ 223

 

Using the values calculated above 221, 200, 223)(221, 200, 223 the average of the values would be calculated below to get the final predicted amount of M&Ms that would fit into the plastic container.

 

\(Mean =\frac{\sum^k_{i=1 }f_ix_i}{n}:n=\sum^k_{i=1}f_i\)

 

\(Mean=\frac{221+200+223}{3}\)

 

Mean = 214.66667

 

≈ 214

 

Therefore, the predicted amount of M&Ms able to fit into the plastic container would be 214.

I bought packets of M&M from the local store and poured it into a plastic container to find the amount of M&MS that actually fit into the container. When filling the container with M&Ms, I did it by ensuring the M&Ms does not exceed the top of the container by using hard cardboard paper and pressing it on the top of the container. I shook the container while pouring in the M&M to ensure the M&Ms fit into the empty spaces and I overfilled the container and then reduced the amount of M&M one by one. After doing 6 trials which each were 225, 221, 219, 219, 221 and 220. I determined the mean amount of M&Ms to be the actual amount of M&Ms that can fit into the container.

 

\(Mean = \frac{\sum^k_{i=1}f_ix_i}{n}:n=\sum^k_{i-1}f_i\)

 

\(Mean=\frac{225+221+219+219+221+220}{6}\)

 

Mean = 220. 83333

 

≈ 220

 

Therefore the average of the actual amount of M&Ms that can fit into the plastic container is 220

In order to examine the accuracy of my investigation, I would be calculating the percentage error of my investigation.

 

\([\text{percentage error }=\frac{accepted\ value-experimental\ value}{accepted\ value}× 100]\)

 

➔ The calculation did would be the experimental value and the actual amount of M&Ms counted manually would be the accepted value

 

\(\text{percentage error} =\frac{220-214}{220}× 100\)

 

percentage error = 2. 727272727%

 

≈ 2. 72727%

 

The percentage error for my investigation as calculated above is 2.72727%, this indicates a minimal error of 2.72727% in calculations and that the investigation was successful. To attain a more minimal percentage error, there are a few possible sources of error that could be addressed. This source of error may come from the approximation that all M&Ms are a perfect elliptical shape as not all M&Ms have a perfect geometrical shape when being manufactured. The imperfections of the M&Ms manufactured may lead to each M&M having a different volume.

 

Overall, this investigation was applicable and useful in several ways in real-world applications. First, I showed the application of mathematical calculations to predict the amount of product that could fit into a container, this is useful for a business when determining the optimum amount of product to maximise the revenue of sales. Furthermore, I have increased my ability to apply mathematical knowledge to problem-solving to enhance my many skills to deduce results to my benefit. Moreover, this investigation expanded my knowledge as a student of integral calculus through the process of learning how to use the volume of revolution and apply it to my investigation. In general, I feel satisfied to be able to complete this mathematical investigation on my own as it proves my mathematical ability does not limit to the classroom; but rather an essential skill that could be applied in the real-world context that would greatly benefit me in the future.

• Abramson, J. (2021, December 21). 8.1 the ellipse - college algebra 2E. OpenStax.
  https://openstax.org/books/college-algebra-2e/pages/8-1-the-ellipse
• Desmos. (n.d.). Graphing calculator. Desmos Graphing Calculator. https://www.desmos.com/calculator
• Donev A, Cisse I, Sachs D, Variano EA, Stillinger FH, Connelly R, Torquato S, Chaikin PM. Improving the density of jammed disordered packings using ellipsoids. https://pubmed.ncbi.nlm.nih.gov/14963324/
• Google. (n.d.). Google drawings - easily create diagrams and charts. Google. Retrieved March 25, 2023, from https://docs.google.com/drawings/d/1WgylmYVsfuvgQSSiRVNtlgVnjckKYcK9T_xFUj KRY5g/edit
• Maverick, J. B. (2023, March 11). Is it more important for a company to lower costs or increase revenue? Investopedia. Retrieved March 28, 2023, from https://www.investopedia.com/ask/answers/122214/company-it-more-important-lower-co sts-or-increase-revenue.asp
• Revision Maths . (n.d.). Volumes of Revolution . Volumes of revolution - mathematics A-level revision. https://revisionmaths.com/advanced-level-maths-revision/pure-maths/calculus/volumes-r evolution
• Stephanie. (2023, March 3). Percent error / percent difference: Definition, examples. Statistics How To. https://www.statisticshowto.com/percent-error-difference/#:~:text=Formula,accepted%20 value
• Weisstein, E. (2023, March 24). Oblate spheroid. from Wolfram MathWorld. Retrieved March 28, 2023, from https://mathworld.wolfram.com/OblateSpheroid.html