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