Surface Area of Metal Cans - Research Paper Example

STUDY ON SURFACE AREA OF METAL CANS Date : Student number- CONTENTS SECTION PAGE NUMBER AIM OF INVESTGATION 3 METHODS 4 ANALYSIS AND RESULTS 5 REFERENCES 11 APPENDICES 12 LOG 13 ROUGH DRAFT 14 AIM OF INVESTIGATION Aim of investigation is summarized as follows: To minimize the surface area of a cylindrical metal can when a fixed volume is given. In the process radius or diameter and the corresponding height of the can are to be determined for such minimum surface area. The analysis will also attempt to find out a general relationship between diameter and height of the cylinder, which will always yield the minimum surface area of the can for any given fixed volume. Finally the analytically found most economic relationship will be compared to the dimensions of commercially available cans with reasoning. An ideal can will be designed with justification. METHODS The problem will be analyzed using following methods: Numerically Graphically By use of excel spread sheets Algebraically By use of calculus ANALYSIS AND RESULTS Part 1 Let us start with fundamental formulas of a right circular cylinder with two circular caps on top and bottom: v = r2h and s= 2r2 +2rh Where, v = Volume of the cylinder r = Radius of the cylinder h = Height of the cylinder s = Surface area of the cylinder with two caps Now let us consider the volumes and surface areas of the following three cylindrical cans: Can 1: r = 3 cm, h = 16 cm. Volume = *32 *16 = 144* cm3. Surface area = 2**32 +2**3*16 =114* cm2 Can 2: r = 4 cm, h = 9 cm. Volume = *42 *9 = 144* cm3. Surface area = 2**42 +2**4*9 =104* cm2 Can 3: r = 6 cm, h = 4 cm. Volume = *62 *4 = 144* cm3. Surface area = 2**62 +2**6*4 =120* cm2 We can see that volumes of all the cans are same but their surface areas are different. Now we can find dimensions of few more cylinders with same volume. It is clear that to keep the volume same as 144* we can factorize the term 144 in different ways to get different cylinders. We know 144 = 48*3 = (4*30.5)2 * 3 and 144 = 72*2 = (6*20.5)2 * 2 Which means we can get two more cylindrical cans of same volume as below: Can 4: r = 4*30.5 cm, h = 3 cm. Volume = *(4*30.5)2 *3 = 144* cm3. Surface area = 2**(4*30.5)2 +2**(4*30.5) *3 =137.5692* cm2 Can 5: r = 6*20.5 cm, h = 2 cm. Volume = * (6*20.5)2*2 = 144* cm3. Surface area = 2** (6*20.5)2 +2** (6*20.5)*3 =194.9116* cm2 Above figures show that Can 2 has the least surface area for the fixed volume given. Now let us display the results in a tabular and graphical form: Can Can 1 Can 2 Can 3 Can 4 Can 5 Surface area (cm2) 358.14 326.72 376.99 432.18 612.33 Height (cm) 16 9 4 3 2 Radius (cm) 3 4 6 6.92 8.48 Ratio of diameter to height 0.375 0.89 3 4.61 8.48 The height Vs radius graph clearly shows that height and radius are inversely proportional for equal volume cylinders. The surface area Vs ratio of diameter to height graph suggests that there is a minimum value of the ratio where the surface area is minimum for a fixed given volume of the cylinder. It also indicates that the value of such ration is close to 1.0 Part 2 In the part 2 we shall try to investigate and find out the exact value of the ratio of diameter to height, which yields the minimum surface area for given volume of the cylinder. Here let us consider that fixed given volume of the cylindrical can is half a liter or 500 cc. Extending the findings of the Part 1, we can probe into the problem in finer detail using the power of Excel spreadsheet. Since we know that the ratio of diameter to height for minimum surface area (for fixed volume) lies close to 1.0, let us study the range of ratio between the values 0.5 to 1.5 in the steps of 0.1 We know v = r2h and s= 2r2 +2rh And for this case v = 500, 2r/h = k where k varies between 0.5 to 1.5 So r = h*k/2 500 =  (h*k/2)2 * h h3 = (500*4)/(*k2) h = ((500*4)/(*k2))1/3 So s = 2**h2 * k2 /4 +2* * h*k* h/2 Or s = *h2 * k2 /2 + * h2*k Or s = *h2 *(k2 /2 + k) Or s = * ((500*4)/(*k2))2/3 *(k2 /2 + k) k s 0.5 366.148 0.6 358.3408 0.7 353.485 0.8 350.6184 0.9 349.1594 1 348.7342 1.1 349.09 1.2 350.0495 1.3 351.4844 1.4 353.2994 1.5 355.422 The above table and the curve more clearly show that the surface area s is minimum when the ratio of diameter to height k is 1.0 The above findings are mainly based on numerical and graphical analysis. Now we shall investigate the problem purely from classical mathematical standpoint so that we can get a generalized solution for any fixed volume of the cylinder. Using v = r2h in the equation s= 2r2 +2rh, we get, s = 2r2 +2r * v * 1/(r2) Or s = 2r2 + 2v/r Or s = 2r2 + v/r + v/r We know from the inequality of arithmetic mean- geometric mean that for positive values of a1, a2, . . ., an, (a1+ a2 +. . .+ an)/n > or = (a1* a2*. . .*an)(1/n) , with equality if and only if a1= a2= . . . = an. Hence (2r2 + v/r + v/r)/3 >= (2r2 * (v/r) * (v/r))1/3 Or s >=3* (2v2) 1/3 which is a constant. So s is larger than or equal to the above constant. To minimize s we shall then consider that s is equal to the constant. That can happen as per above mentioned theorem only when 2r2 = v/r, or 2r2 = r2h /r, or 2r = h, or d = h So now we got a mathematical proof that surface area of a cylindrical can is minimum for a given fixed volume when diameter to height ratio is equal to 1. We shall now try to establish our results using principles of calculus. Earlier we have already derived the equation of surface area in terms volume and radius as below: s = 2r2 + v/r + v/r Now differentiating both sides with respect to r we get, ds/dr = 4r – 2v/(r2) For minimum or maximum value of s, we set ds/dr = 0 = 4r – 2v/(r2) Or 4r = 2v/(r2), or 4r = 2r2h /(r2) Or 4r3 - 2r2h = 0 Or 2r2 *(2r-h) = 0 Hence either r = 0 or 2r = h. Since r = 0 has no physical significance, we shall consider 2r = h, or d = h Now to see whether d = h yields minimum or maximum value we shall have to see that (d2s)/(dr2) = positive or negative. (d2s)/(dr2) = 4 + 4v/(r3) Since the above expression is always positive for all positive values of v and r, we can conclude that for d = h the surface area s is minimum for any given fixed volume v. Hence it is now established that for any fixed volume of a cylinder the ratio of diameter to height shall always be 1.0 when the surface area with two ends capped is minimum. For a 500 cc can then we can determine the dimensions of the can using the relationship of d = h to get minimum surface area, 500 =  * (d/2)2 * h =  * (h/2)2 * h =  * h3 *(1/4) Or h = ((500*4)/)1/3 or h = 8.6025 cm and d = h = 8.6025 cm Part 3 The table below shows the dimensions of few cans commercially used: PRODUCT AND SIZE DIAMETER (D) HEIGHT (H) RATIO (D/H) Baked beans, 425g 7.65cm 10.5cm 0.714 Pepsi can, 355cc 6.6cm 12cm 0.55 Coke can, 12oz 6.6cm 12.1cm 0.545 The above sample data shows that ratio of diameter to height used in commercial world is not always very close to the optimum ratio of 1.0 which yields the use of minimum metal for cans. Then the question is why the manufacturers do not always go by the mathematically proven optimal ratio of diameter to height to minimize their cost of metal for cans? Some of the answers lie in the following considerations: The can diameter must be suitable for easy handgrip for say schoolboy to aged person. The actual construction of cans in most cases is not exactly like cylinders, especially near bottom where it is dome shaped and in top its diameter is somewhat less. This shape does not allow mathematically deduced optimal ration formula to be applied fully for commercial cans in many cases. This situation arises because for economy and ease of construction, exact cylindrical shapes do not become a feasible solution in most real life situations. Packaging consideration dictates shape to some extent. Sometimes a different shape is purposely adopted by manufactures for unique branding even at the cost of more metal. Part 4 Hypothetically if I have to design the ideal shape of a cylindrical soda can of 355ml, I shall consider that the diameter to the height ratio must be as close as possible to the most economic ratio of 1.0 giving due respect to the basic requirements of easy gripping, ease of construction and packaging convenience. Earlier in our mathematical analysis we have seen that surface area increases sharply as soon as we start deviating from the optimal ratio of diameter to height. Now commercially available soda cans are of 355ml are of about 6.6cm in diameter. I have checked that diameter of up to 7cm can be adopted for good gripping by average persons. So my choice will be to adopt the 7cm diameter. Then we get 355 = *(7/2)2 * h Or h = 9.224cm And then s = 2*(7/2)2 +2*(7/2)*9.224 = 279.815 sqcm Now for 6.6cm diameter for commercially available cans, we get the following relation thinking hypothetically the shape as perfect cylinder: 355 = *(6.6/2)2 * h Or h = 10.376cm And then s = 2*(6.6/2)2 +2*(6.6/2)*10.376 = 283.565 sqcm The percentage of metal saving with 7cm diameter compared to 6.6 cm diameter then works out to: (283.565-279.815)*(100/283.565) = 1.322% Since soda cans are produced in millions the savings of 1.33% would be substantial. To alleviate any problem of gripping by children in case of 7cm diameter, I would prefer to introduce another 200ml can for children with lesser diameter. This will be welcome for children and parents since youngsters most of the time do not drink 355ml at one time. Here I would choose 5.5cm diameter and find the height as follows: 200 = *(5.5/2)2 * h Or h = 8.418cm REFERENCES 1. Text book of Calculus 2. Text book of Algebra 3. Text book of Geometry 4. Internet sites APPENDICES Not used LOG Step 1: Study of reference textbooks and understanding of basics. Step 2: Finalization of methods of analysis an investigation required. Step 3: Actual analysis work Step 4: Review of results and modifications required in analysis work Step 5: Collection of data for commercially available cans Step 6: Discussion with fellow students and teachers about the present practices and improvement in can dimensioning. Step 7: Internet surfing for relevant details as available. Step 8: Decision making about the dimensions of ideal hypothetical can ROUGH DRAFT Read More