Mathematics the cut-out corner has a fixed

 

 

 

 

 

 

 

 

 

 

 

 

Mathematics SL – Internal Assessment

 

Topic: Finding the optimal value of a volume of a box using differential calculus.

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Introduction:

While I was at the DHL office to send something to my host family, the man, who works there, gave me a square-based box to put the stuff into. The box was small and not all my stuff fit in there, so I asked for a larger one, which was a rectangular-based box. After being introduced to optimisation, which is the process to find the value of maximizing or minimizing a function (Haese, 2012), I found that the box from the DHL office should have been made according to a mathematical formula or mathematical method to give a maximum volume for its contents.

After researching, I discovered that the boxes were made by cutting out the corners of the carton used to make the box. I want to find out if the cut-out corner has a fixed value of length. How do these cut-out corners used to form a box with a maximum volume, change as one side of the original carton is increasing in size? Also, if I have a fixed surface area, How can I find the maximum volume of the box?

In order to reach an answer to my questions, I will use the process of optimisation of one or more variables. To do that I will be using different approaches, such as technologically, analytics and differential calculus to obtain exact solutions.

 

How do the cut-out corners used to form a box with a maximum volume, change as one side of the original carton is increasing in size?

To answer this question, I start with a square because from it I can make the smallest box. I choose it to be my basic shape that I will base the results I will have to it. I decided to find the relationship between the length of the side of the original square and the length of the cut-out corners.

I used a square that measures a units by a units. The cut-out corner measures x units by x units – as shown in the figure 1.

x

a – 2x

Let :

, where x is in the interval 0, :

Finding the first derivative for the volume:

t n val 0, in teh  tes bz a ununites bz a unt out from each corner.

Let :

 

Figure 1: Square a by a units.

I see that it has factors, so:

 

 

Since we have two values, I will try which value of them is yielding me to a maximum volume of the box formed from this square. I will use the second derivative to see that:

As seen from the previous results, when , the value of the volume will be greater than the value when . Therefore, it is a minimum. This solution made me realize that x is the variable of differentiation, and a is constant with the regards to differentiation.

So, I made a general result to this, which says that: to find a box with maximum volume, we can start with cutting the corners with a length equals to  the length of the original square.

x

a – 2x

b – 2x

Now, I will start to increase one side of the square.

In figure 2, the side b is increasing in size. I want to find the value of x for any values of a and b which allows to have a maximum volume of the box.

Figure 2: Increasing the length of the side.

Let :

, where x is in the interval 0, :

                    

Finding the first derivative for the volume:

t n val 0, in teh  tes bz a ununites bz a unt out from each corner.

Let :

Solving by using the quadratic formula:

For any value of b ? a, the positive root solution will be greater than or equal to , so it will not give me the maximum solution I am looking for since it is outside the possible values for a. Therefore, the negative root solution will give me the maximum, which is the following:

This equation can help me solve the box problem I have. When a = b, it will give me the solution I found when I maximise the volume of the square measures a units by a units.

To have an idea of the solution to my question, I decided to assume one of the sides as a fixed value. Since b is the one which is increasing, I assumed the value of a as fixed one. I choose 6 units equals to a to make the calculation easier and to have a clear result. Then, I considered f as a function of only b. the result is:

I used the calculator to make the graph of the function f(b). I used the TI-Nspire™ CX to graph the function. I assumed that the role of f(b) will be represented by y, and the role of b will be represented by x. Therefore, the x-axis represents b which is the length if the rectangular, and the y-axis represents x which is the cut-out corner’s size.

So:

Figure 3, represents the function of f(b).

 

Figure 3: The function of f(b).

Since the function is in both negative and positive areas of the graph, I decided to only take the positive side of the function as the negative side does not make sense in the real life. So, I take the zoom into x: 0, 100 and y: 0, 3. This is represented in figure 4.

Figure 4: The function of f(b) after the zoom.

I notice from figure 4 that as the size of b increases beyond 6, the cut-out size x, for the box with maximum volume also increases; but it seems that there is a limiting value, that is, a horizontal asymptote for the graph does seem to exist.

 

 

I used the table feature on the calculator, and tried to see higher values for b. figure 5 shows the result. I found out that the limit for b seems to be around 1.5 units.

 

 

 

Figure 5: The table feature for the function Y.

 

To make sure, I represent a new function:

The function was so close to the function f(b), but does not have any point of intersection with it. This means that there is a limit for the value of x and this value is around 1.5 units.

 

a

b

y

6

57338

1,499960757

1

57338

1,49999346

2

57338

1,499986919

3

57338

1,499980379

4

57338

1,499973838

5

57338

1,499967298

7

57338

1,499954216

8

57338

1,499947675

9

57338

1,499941134

10

57338

1,499934593

20

57338

1,499869174

30

57338

1,499803744

40

57338

1,499738302

50

57338

1,499672849

60

57338

1,499607385

70

57338

1,499541909

80

57338

1,499476422

90

57338

1,499410923

100

57338

1,499345413

1000

57338

1,493402684

Figure 6: The new function which is

 

 

This result seems really interesting, so I decided to try a couple of other values for a and I can see my result in a more general pattern. I used Excel to do that. 

Using Excel, I put in different values of a, and I decided to keep b constant and equals to 57338, which is the value of b which is represented in the table feature for the function y in figure 5. y is representing the following function:

The results I got are represented in table 1. I found out that the value of y is always coming closer to 1.5.

So therefore, in a general level: the value of x always seems to approach , where a is the fixed side of the square.

Table 1: The results of the Excel sheet.

 

 

This result made me more excited and I decided to double check it and use the limit to evaluate it:

I will evaluate this limit by rationalizing the numerator, this is as the following:

Since the limit of the sum equals to the sum of the limit, and b         ?:

So, this value matches the results I got using the calculator and the Excel sheet, which made me more confident on my results.

In conclusion and as an answer to my question, I can say that for any size of a units by b units, where b ? a, and the cut-out corner size, x, will yield the box to the maximum volume as:

The square uses the smallest cut-out size (when ); and the more extended the side will be, the closer the cut-out size value should be to .

What is the maximum volume of a box, which has a fixed surface area?

To answer this question, I decided to have a fixed surface area, which will help me more in finding the maximum volume of the box. I assumed that my fixed area for the box is equal to 70cm3, this is to have easier calculations and clearer results. I also choose a box measured by h units by w units by l units as shown in the figure 7.    I will maximize the volume of the box using the methods of maximizing two variable functions.

Figure 7: The box of h unites by w unites by l unite.

 

And since I assumed that the area is 70cm3:

I will solve for h in terms of w and l:

       

       

Now, I will write the volume in terms of w and l:

I will maximize the volume for positive values of l and w; since the negative values does not make sense in the real life for physical reasons. I will find the points of V where l ? 0 and w ? 0. After that I will test these points and find the maximum volume.

The volume function is consisting of 2 variables, and I will maximise the function using the methods of functions of two variables. My aim is to find the maximum of the function of h = V(l, w), where l and w are my independent variables and h is the dependent variable. The graph of the function is a surface in three-dimensional space, this graph might have maximum points or minimum points (or might both). But for the surfaces there is a third possibility – a saddle point, this point is also called as stationary point, which is a maximum or a minimum point (Wicks, 2011).

 

 

 

I will find the first derivative for the point h = V(l, w). I will do this by finding the first derivative of each variable alone in respect of the other as the following:

Ø  The first derivative of l in respect of w:

Ø  The first derivative of w in respect of l:

 

Therefore, the point is:

 

Now, in the real life, there cannot be a physical length which is equal to 0cm. So, I assumed that l and w are nonzero, which will yield me to solve the following equations to find the value of w and l which will represent to me the stationary points:

And to solve for it, I need to subtract the second equation from the first one:

Since l and w are both positive because of the physical – real life – reasons. So:

This result is really interesting, since l and w are the same in size, that means that the box has a square base. 

Now, I can find the value of l by adding its value I found into the equation. This will be as following:

The negative result will be ignored because of the physical – real life – reasons. So:

Now, let’s put everything together again:

I found out that l = w and  . so, we have the following point:

This means that the value of l = w = h , which is really interesting to see. And this also means that the box is a cubic box.

 

 

 

 

 

 

 

Now, the boundary of this domain are the lines   and  for w, l ? 0. Remarking that on those lines, both of the surface area and the volume will be zero, I will disregard those because of the unrealistic physically solutions. Thus, my critical point is the only test point. The volume at this point is:

BUT, is this volume the maximum one?

I can answer this by considering two observations, which makes me sure that it is the maximum:

1.       For a fixed l (assuming that l = 1), we have :

·         As w            0, we see that h            35 and the volume tends to be zero.

2.       As l and w tend towards ?, we can see that h eventually becomes negative, and this is an unreasonable physical assumption.

To make sure and support the results I got, I decided to use technology to see if the points and the volume I got are right. So, I used WolfarmAlpha©1 website to help me checking my results and graphing them as well.

I plotted the following function of the volume:

I made l is equal to x and w is equal to y, so that I will have the x-axis for l and y-axis for w and z-axis for the Volume. The website will give me the maximum volume and will tell me at which points this volume is happening. So, I entered the following function to find its maximum volume:

 

I got the following results:

 

 

 

Figure 8: The local maximum of the Volume function.
Source:http://www.wolframalpha.com/input/?i=maximumof+%5B%2F%2Fmath:(35xy-x%5E2.y%5E2)%2F(x%2By)%2F%2F%5D

 

 

 

 

 

 

 

 

 

 

 

Figure 9: The 3D plot of the Volume function.
Source:http://www.wolframalpha.com/input/?i=maximumof+%5B%2F%2Fmath:(35xy-x%5E2.y%5E2)%2F(x%2By)%2F%2F%5D

 

 

 

 

So, figure 8 represent the local maximum for the function, which is happening at the point (3.41565, 3.41565) these are the same values I got for this function, also, the maximum for the function is the same in both my results and the website. From figures 8 and 9, I can see that these results from the website supports the results I got.

Therefore, I conclude that the critical point is a maximum. Thus, the ultimate volume is happening at the point:

Where the volume is:

So, that means that the maximum volume of a cuboid is a cube. And for a fixed surface area, I can get a maximum volume by using a cubic box.

 

It would be really interesting to find other problems related to optimisation and trying to solve these problems. This will allow us to be more efficient and sustainable, since this will help us to know the lower amount of materials we need to use in order to get the maximum volume for a fixed surface area.

 

 

 

Bibliography:

·         “Alpha: Making the World’s Knowledge Computable.” Wolfram, www.wolframalpha.com/input/?i=maximumof%2B%5B%2F%2Fmath%3A%2835xy-x%5E2.y%5E2%29%2F%28x%2By%29%2F%2F%5D.

Accessed 25 Nov. 2017.

 

·         Haese, Robert, et al. “Chapter 17: Applications of Differential Calculus – C: Optimisation.” Mathematics SL for the International Student (IB Diploma Programme), Haese Mathematics, 2012, pp. 428–431.

Accessed 25 Nov. 2017.

 

·         Wicks, Mark A. • Max/Min for Functions of Two Variables. 2011, personal.maths.surrey.ac.uk/st/S.Zelik/teach/calculus/max_min_2var.pdf+.

Accessed 25 Nov. 2017.

 

 

1 http://www.wolframalpha.com/