PROJECT A

Table of Contents1. Project Title22. The Problem Statement33. Mathematical Formulation43.1 The important elements of the problem and relationships among elements43.2 Mathematical Concepts and Principles53.3 Model of the problem mathematically63.4 Mathematical Model/ Methods Chosen74. The Problem Solution94.1 Assumptions94.2 Solution of Problem9i.To determine the minimum radius of the cylindrical can.9ii.To determine the minimum surface area for the cylindrical can12iii.Proof for the minimum radius of cylindrical can13iv.To determine the minimum height of the cylindrical can14v.Proof for the relationship between the height and the radius of the cylindrical can144.3 Graphical Representation of the Problem Solution15i.Diagram of a cylindrical can of radius r and height h.15ii.Diagram showing a picture of the solution obtained by Minimizing the Surface Area of a Cylinder with a Fixed Volume15iii.Graph 1 Showing the Surface Area of the cylindrical can as a function of the cans Radius.16iv.Graph 2 - Showing the Surface Area of the cylindrical can as a function of the cans Height.175. Application of Solution185.1 Consideration of solution and its application to the given real world problem185.2 Discussion on the solution of proof to the given problem196. Conclusion24

1. Project Title

Optimizing the surface area of a commercial cylinder with a given fixed volume.

2. The Problem Statement

The problem being solved for industry is to increase the size of a can of soda (case being studied is Coke soda), while minimizing the amount of tin used to make the container. This analysis seeks to increase the height of the can of soda over its current height without using additional tin material and maintaining the fixed volume. Generally the height of a 12 ounce can of soda is approximately 12. 5 centimeters and its radius is about 2.75 centimeters. The constraints of the project includes: Maintaining the current volume of 12 ounces of soda Minimizing the amount of tin material used to make the can by determining the minimum surface area of the can for a fixed volume. The new can design should appear as if more coke is being sold to the customer by enlarging it. Hence, determine the new radius and height of the can for a fixed volume and minimum surface area.

Decisions to be derived are:i. To establish the relationship between the radius and the height of the can in order to minimize the surface area for a fixed volume.ii. To determine the new shape of the 12 ounce can of soda that can minimize the amount of aluminum used in the can. This includes:a. The new height of the can.b. The new radius of the can.

3. Mathematical Formulation

3.1 The important elements of the problem and relationships among elements

The shape of the can of soda is a right circular cylinder, which is tall and slender in shape with a fixed volume as shown in figure 1.

Figure 1: Cross-Sectional and Transverse Section of a Cylindrical Can

Where: - The area of the base of the cylindrical can + the area of the top of the cylindrical can. - The surface area of the lateral surface. - The total surface area of the cylindrical can. - The volume of the cylindrical can. - The radius of the cylindrical can. - The diameter of the cylindrical can. - The height of the cylindrical can. 3.2 Mathematical Concepts and Principles

The surface area (S) of the cylindrical can = {the area of the base + the area of the top} + the area of the lateral surface

S =Where: - The area of the base of the cylindrical can + the area of the top of the cylindrical can. - The area of the lateral surface.

Therefore, the surface area (S) of the cylinder may be represented by the following:S =Thus

The volume V of the cylindrical can is fixed. = The volume of the cylindrical can. = The Circular area * the height of the cylindrical can.

3.3 Model of the problem mathematically

1. What are the dimensions of the coke can with a 12 ounce capacity that uses the least amount of tin?a. We need to find the minimum surface area of the can for a fixed volume. b. The unknowns to determine the minimum surface area of the can will be the radius and the height.c. The volume is fixed as the can holds 12 ounces.d. The surface area of the cylindrical can, can be calculated and the volume of the can, can be calculated. However one of the variables (radius or height) must be removed from the equation, so that it can be solved.2. What is the relationship between the radius of the cylindrical can and the height of the cylinder in order to minimize the surface area for a fixed volume?

3.4 Mathematical Model/ Methods Chosen

In determining the minimum surface area for a can of a fixed volume, we will use the method of applying the derivative to the equation to find the minimum value of the surface area as a function of the radius. The minimum occurs at a point where the derivative of the function (in this case, the derivative of the surface area) is equal to zero.

The analysis should conclude that the first derivative of the equation for surface area when equal to zero should produce a curve which has a horizontal tangent line and at this point, if it is a minimum value, then the curve would have the shape of a valley.

Therefore, the process of calculation to determine the minimum radius and the minimum surface area required for a volume of 12 ounces of soda will be done as follows:i. Identify the equations to be used to determine the minimum surface area of the cylindrical can. These are:a. Surface area of a cylindrical can.b. Volume of a cylindrical can.ii. Reduce the amount of variables in the equations to an equation with one variable. The variables are:a. Radiusb. Heightiii. Find the First Derivative of the equation to determine the slope.iv. The minimum surface area occurs when the slope is zero. Therefore, equate the first derivative to zero. v. Solve for the minimum radius of the can.vi. Run a test to determine whether the value is a minimum or a maximum by carrying out the Second Derivative test.vii. If the Second Derivative is greater than zero (positive) then it proves that the critical point obtained from the first derivative is the minimum point. (minimum radius for a minimum surface area)viii. Determine from the results the relationship between the radius and the height in order to minimize the surface area of a can of fixed volume. Calculate the minimum radius, the minimum surface area and the height.ix. Use the values for radius to plot a graph to show a. the surface area of a cylindrical can of fixed volume as a function of the radius and,b. The surface area of a cylindrical can of fixed volume as a function of the height.

4. The Problem Solution

The Problem Solution aim is to determine the minimum surface area (S) of the cylindrical can of volume V with r being the radius of the cylinder and h being the height of the cylinder. The surface area of the cylinder is represented by S.

4.1 Assumptions

Assume that the volume is fixed and constant. The fixed volume for the can is 12 ounces, which is equivalent to 354.88

The shape of the can will be determined by the minimum surface area of the can to ensure that the minimum amount of material is used to make the can. Therefore it is important to determine the relationship between the radius and the height of the can in order to minimize the surface area of the can for a fixed volume.

4.2 Solution of Problem

i. To determine the minimum radius of the cylindrical can.

The surface area (S) of the can:

The cylindrical can holds a fixed volume (V) of approximately 354.88

The volume of a cylindrical can (V) is:

354.88

Therefore, Making (h) the subject of the formula

To determine the minimum surface area, Substitute (h) in the surface area formula to obtain one independent variable.

By simplification,

To determine the minimum surface area of the cylinder, we will differentiate the Surface Area (S) with respect to the radius (r).

Differentiating we get:

Similarly:

The minimum surface area of the can is determine when the slope is equal to zero. This minimum value will provide the minimum radius for the can where the minimum surface area is achieved.

Thus, take the derivative of the Surface area and equate it to zero.Let:

Solving for (r) we get:

Cube rooting both sides we get:

Minimum Radius (r) of Cylindrical Can

ii. To determine the minimum surface area for the cylindrical can

We shall substitute the value of radius (r) obtained above into the equation for Surface Area. That is

Substituting r = 3.84,

Therefore the minimum surface area for the cylindrical can of a fixed volume, 354.88 has a Surface Area (S) of 277.43

iii. Proof for the minimum radius of cylindrical can

In order to determine if the radius found above is a true minimum of the cylindrical can, we must take the second derivative test of the surface area. From the calculations above, the first derivative is:

The second derivative is:

Substituting the minimum value of the radius r = 3.84 we get:

Because the radius is positive, the second derivative is positive and the critical point we obtained initially from the first derivative of the surface area is called the minimum point or the minimum radius. Hence the minimum radius is 3.84 for a cylindrical can with a fixed volume of 354.88. The minimum surface area of the can is 277.43

iv. To determine the minimum height of the cylindrical can

We will substitute the minimum radius (r = 3.84 cm) into the equation:

Where, 354.88 Therefore the minimum height (h) is,

v. Proof for the relationship between the height and the radius of the cylindrical can

The above value of the height of the cylindrical can proves that the minimum surface area for a fixed volume occurs when the height (h) is two times the radius (r).

Substituting r = 3.84cm we get:

7.68

4.3 Graphical Representation of the Problem Solution

i. Diagram of a cylindrical can of radius r and height h.

ii. Diagram showing a picture of the solution obtained by Minimizing the Surface Area of a Cylinder with a Fixed Volume

iii. Graph 1 Showing the Surface Area of the cylindrical can as a function of the cans Radius.

iv. Graph 2 - Showing the Surface Area of the cylindrical can as a function of the cans Height.

5. Application of Solution

5.1 Consideration of solution and its application to the given real world problem

We have determined that the ratio of the height (h) of the right cylindrical can to the radius (r) is when h = 2r. That means the optimal solution occurs when the height is equal to twice the radius or the height is equal to its diameter. In consideration of the solution achieved the following will be considered:

i. Compare the dimensions of the current can of soda with the optimized cylindrical cans dimensions. Determine how this relationship between the height and the radius compare to the cylindrical soda cans in the supermarket, namely the coke can? ii. Determine the optimal height for the new design, using the same amount of aluminum material?iii. What is the impact on the manufacturing of the new can?

5.2 Discussion on the solution of proof to the given problem

The solution obtained in section 4 shows that for a right cylindrical can with a fixed volume of 354.88 the minimum surface area of the can obtained by optimization of the can to use the least material is 277.43 The optimization of the can was achieved for a minimum radius of 3.84 and a height (h) of 7.68 cm.This means that the minimum surface area achievable for a fixed volume occurs when the height (h) is twice the radius (r), or when the height equals the diameter (d).

i. How does this relationship between the height and the radius compare to the cylindrical soda cans in the supermarket, namely the coke can?

Table 1 - shows the comparison of the dimensions for the Actual Cylindrical Can in the supermarket with the Optimized Cylindrical Can:

DimensionsActual Coke Can in SupermarketOptimized Can based on Analysis

Volume12 ounces12 ounces

Radius2.75 cm3.84 cm

Diameter5.5 cm7.68 cm

Height12.5 cm7.68 cm

Surface Area263.3 277.43

The restriction imposed in order to optimize the radius and height of the cylindrical can was on the volume of the can. The volume maintained was 12 ounces or 354.88 . This will be approximated as 355 ml.

The above measurements show that the actual cylindrical cans dimensions are different from the dimensions obtained by the optimization technique. Hence, as the dimensions are not similar, this proves that the cylindrical can in the supermarket has not been optimized.

The diagram in Section 4.3.ii, as shown below, shows that the optimized results gives a can with a very short and wide shape. This is not the shape of the actual can.

This shape would not be ergonomically suitable for the customer and would put additional constraints on manufacturing of the can.In considering the actual cylindrical can we identify the physical dimensions on the Graph 3 below. This shows that the manufacturer considered the ergonomics of the can after the optimization test were done and adjusted the height to produce a can that would be easier to hold and also package.

Graph 3 - Showing the representation of the actual soda can dimensions on the graph of Surface area as a function of height.

ii. Determine the optimal height for a new design, using the same amount of aluminum material?

Graph 4 shows the Surface Area as a function of height for the new dimensions of the can design. This can design would be slender in shape, of approximately 16. 7 cm in height with a radius of 2.5 cm. This would be quite similar to the shape of the Red Bull can, which has a height of 13.5 cm and radius of 5.1cm. However the coke soda can has a volume of 12 ounces (355ml), while the Red Bull can has a volume of 8.4 ounces (250 ml). Hence the coke can would need to be taller to accommodate the volume of 355 ml.

iii. Consider the impact on the manufacturing of a new can with new dimensions.

The design of a new coke can would first of all draw a lot of attention from the customer. The can would be more slender, easier to hold and carry than the original design. The can would appear as if the customer was actually getting a can with more volume because of the increased height. I believe the customer would welcome a change as long as the formulation of the product, that is the coke soda in the can taste the same as before.

From a manufacturers point, however it has been shown that Coke soda has not modified their can since the modification done in the 1980s. They have not tried the Red Bull approach of giving the customer a slender, new look can. The manufacturers may have restricted themselves with development because a new design would mean that new machinery would be required, even if the same volume and minimum surface area for the aluminum material was used.

It would cost an additional amount of money to change the machinery that produces the can. They would only consider redesigning the can if it would generate additional revenue to cover this cost, as well as produce more sales than it is currently producing. Hence a market survey would have to be done on a sample can with the new design dimensions to see how feasible redesign of the shape of the can would be and customers perception of the new shape.

6. Conclusion

The investigation using calculus method of differentiation to determine the optimal height of a right cylindrical can, which uses the least amount of material and contains a fixed volume of 12 ounces (355ml) has shown that a minimum surface area is obtained when the height is equal to twice the radius.

However the results obtained shows that the actual can did not use these optimized results in the design dimensions of the can. The optimized can had a shape that was shorter and fuller, which would not be easily held by the customer. The actual cans dimensions were taller with a smaller radius. The actual can was also narrowed at the top and bottom. This was not considered in the analysis. Both cans however maintained the same volume.

The analysis proved that the value of the radius (r) obtained after differentiation gives a minimum surface area for the cylindrical can. The height of the can obtained however of 7.68 cm is very different to the actual can. This shows a difference between the optimized results and the actual dimensions of the can. The reasoning behind the actual cans taller dimensions over the using the optimized dimensions, would be that the shape of the optimized can would not be ergonomically viable for a consumable product. It would be too wide, too short, and the customer could not drink the soda from the can easily.

The optimized dimensions and shape would also have an impact on packaging and transportation of the product. The optimized can could not hold in the palm of the hand of the consumer, it would also require cartons with larger dimensions to package the product. Transportation costs would also increase due to the wider product utilizing more packaging space and therefore more volume per shipment.

The actual cans dimensions would also suggest that the manufacturer did not use the minimum of material to make the can, because the dimensions for height and radius are different from the optimized values.

I believe that the manufacturers of the coke soda can, could consider redesigning the shape of the can. As the world becomes more environmentally conscious, and many health factors are at the forefront of manufacturing, they could consider a slender, taller can, with similar volume of 12 ounces. However to boost sales and customers acceptance of the new design product, they would need to create a new more healthy formulation of a soda with less sugar and additives.

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