• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28

Math portfolio: Modeling a functional building The task is to design a roof structure for the given building. The building has a rectangular base 150 meters long and 72 meters wide. The height of the building should not exceed 75% of its width

Extracts from this document...

Introduction

Math portfolio: Modeling a functional building

The task is to design a roof structure for the given building. The building has a rectangular base 150 meters long and 72 meters wide. The height of the building should not exceed 75% of its width for stability or be less than half the width for aesthetic purpose. The minimum height of a room in a public building is 2.5 meters.

The height of the structure ranges from 36m to 54 (72×75%) m as per the specification.

At first I will model a curved roof structure using the minimum height of the structure that is 36 meters.

From the diagram given, the curve roof structure seems to be a parabola hence I will use a general equation of parabola that is

y= ax2 + bx +c  -----------(1)

Now the width of the structure is 72 meters and the height is 36 meters.

Let the coordinate of the left bottom corner of the base is (0,0)

Then the coordinate of the right bottom corner will be (72,0) and the coordinate of the vertex of the parabola will be (36,36)

Since the above three points lies in the parabola, we will get 3 equation by substituting these coordinate in equation (1)

C=0  -------(2)

5184a + 72b = 0  -------(3)

1296a + 36b = 36  ------------(4)

Solving:

5184a + 72b = 0

5184a =-72b

a =image23.pngimage23.png

a = image55.pngimage55.png

Substituting a = image55.pngimage55.png to (4)

1296(image55.pngimage55.png ) + 36b = 36

-18b + 36b = 36

b = 2

Therefore: a =image24.pngimage24.png , b = 2  , c = 0

So the equation for the curved roof structure of 36 meters height will be

y = image24.pngimage24.pngx2 + 2x  -----(5)

The given equation is graphed as shown below.

image31.png

Now we need to find the dimensions of the largest possible cuboids which would fit inside the above curved roof structure.

I know that the length of the cuboid is 150 meters. So I have to find out the height and the width of the cuboid.

...read more.

Middle

 -image25.pngimage25.png)-0]

= 48H

Volume of the structure = 150×48H=7200H

Using this I will be able to calculate the ratio of wasted space to the volume of the office blocks for the structure height ranging from 36m to 54m.

image26.jpgThe above table shows that the ratio of the wasted space to the office block is the same for different height of the structure.

I will calculate the total maximum office floor area in the block for different heights with the given specification.

Minimum height of a floor = 2.5meters

We know that h=0.67h

Area of the floor =width × height

=41.56×150

=6234m2

Total maximum floor area=number of floors × 6234m2

image27.jpg

The above table shows that the number of floors that can be constructed in a building increase with the increase in the height of the structure. The total maximum office floor area increase with the increase with the number of floors that can be constructed

Now I will do the whole investigation when the façade is placed on the longer side of the base.

So the width of the base of the rectangular building = 150m

And the length of the rectangular building = 72m

Here height of the structure varies from 50%of 150m to 75% of 150mwhich is from 75m to 112.5m

Let the coordinates of the left bottom corner of the base is (0, 0)

Then the coordinate of the left bottom corner be (150, 0),and the coordinate of the top corner be (75, 75)

Since the above 3 points lies in the parabola, we will get 3 equations from equation(1)

C=0

5625a + 75b= 75

→75a + b =1  ----(14)

22500a + 150b = 0

→ 150a + b = 0  ----(15)

Solving:

(15)-(14):

75a = -1

a =image28.pngimage28.png

Substituted a =image28.pngimage28.png to (14):

75(image28.pngimage28.png ) + b =1

-1 +b =1

b = 2

We get a =image28.pngimage28.png , b=2, c= 0

So the equation for the curved roof structure of 75m height will be

Y=image28.pngimage28.pngx2 + 2x ------(16)

The above equation is graphed below.

image29.png

...read more.

Conclusion

Volume of the cuboid=150×55×2.5=20625m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5=17.5

y = image24.pngimage24.pngx2 + 2x

17.5= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -17.5=0

x2-72x+630=0

x=10.19, 61.81

Width of the cuboids base= 61.81-10.19=51.62m

Volume of the cuboid=150×51.62×2.5=19357.5m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=20

y = image24.pngimage24.pngx2 + 2x

20= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -20=0

x2-72x+720=0

x=12, 60

Width of the cuboids base= 60-12=48m

Volume of the cuboid=150×48×2.5=18000m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=22.5

y = image24.pngimage24.pngx2 + 2x

22.5= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -22.5=0

x2-72x+810=0

x=13.95, 58.05

Width of the cuboids base= 58.05-13.95=44.1m

Volume of the cuboid=150×44.1×2.5=16537.5m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=25

y = image24.pngimage24.pngx2 + 2x

25= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -25=0

x2-72x+900=0

x=16.1, 55.9

Width of the cuboids base= 55.9-16.1=39.8m

Volume of the cuboid=150×39.8×2.5=14925m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=27.5

y = image24.pngimage24.pngx2 + 2x

27.5= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -27.5=0

x2-72x+990=0

x=18.51, 53.49

Width of the cuboids base= 53.49-18.51=34.98m

Volume of the cuboid=150×34.98×2.5=13117.5m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=30

y = image24.pngimage24.pngx2 + 2x

30= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -30=0

x2-72x+1080=0

x=21.30, 50.70

Width of the cuboids base= 50.7-21.3=29.4m

Volume of the cuboid=150×29.4×2.5=11025m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=32.5

y = image24.pngimage24.pngx2 + 2x

32.5= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -32.5=0

x2-72x+1170=0

x=24.78, 47.22

Width of the cuboids base= 47.22-24.78=22.44m

Volume of the cuboid=150×22.44×2.5=8415m3

The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5+2.5=35

y = image24.pngimage24.pngx2 + 2x

35= image24.pngimage24.pngx2 + 2x

image24.pngimage24.pngx2 + 2x -35=0

x2-72x+1260=0

x=30, 42

Width of the cuboids base= 42-30=12m

Volume of the cuboid=150×12×2.5=4500m3

THE GRAPH WHICH CONTAIN DIFFERENT 19 CUBOID

image59.pngimage11.pngimage10.pngimage09.pngimage03.pngimage21.pngimage04.pngimage14.pngimage22.pngimage13.pngimage19.pngimage20.pngimage02.pngimage07.pngimage18.pngimage08.pngimage05.pngimage15.pngimage06.png

Total volume of the cuboids are=246386m3

For 36m height, the volume of the structure= 7200h = 7200×36=259200m3

Wasted space= 259200-246386=12814m3

So the ratio of the volume of the wasted space to the office block is image60.pngimage60.png

The ratio of the volume of the wasted space to the office block is 0.72 in case of single cubiod arrangement whereas there is only 0.05 in the case of multiple cuboids. From the result we can say that for using multiple cuboid is more economical!!

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    C q8�S�`���?�z����)�.}� ���Y"%a"��W��x',b'LC0"��i"-8ã¯ÎH�{��(c)S���>����g�K[-a3������" +~�(r)�� X��Q7W�'a�3/4�3/4�.��'k(r)�/�7�tS�;U\G| �o��L��a m��g'...�+�g[�3/4(c)$lRh(i�-�P��E�-�t�Kx [h�"�g6M>�-�ĺF�1/4W�UN2*V�߿ß�@C�"�=�]1/4�{�|�2�k�-���tuD롹�}2|�\�;"�3���_�"�d$d׬4D�%Or�gK *����(c)`qe��]�<;T�ٱ�7QЧ�;��5-�B"�!� PbØ!\� ���b���@s7>'����+"��@���+�l(b�#�46�(c)�c�'Y�+o(tm)/���5�H�pN'Yq��T�.I�1/4-���-�sD"F �-�U�<(tm)'�@�+�C�� �L E�( ������D��r�#"\Jӧ���� )ËtfD����*K�3/4""�[email protected]Ê ï¿½ï¿½ï¿½ï¿½)�NÙ@��S�1͡� v @l�� " )6�uG1/2�v���I�@@-� Up�å­# -e"�8-�>��yD�!�b���IEND(r)B`�PK !x�4�4word/media/image8.png�PNG IHDRT'�"�RiCCPICC Profilex�YwTͲ�(tm)�,KXr�9�s"�a�K�Q��"� H ���*��""(�[email protected]�PD$1/2A?�{�y��ys����v���� �JTT�@xD\��(tm)!��" ?n�:� ����F��Y#��\����5&�#�0�72�@��������k��~Q1q ~ �'ĸ(� ��(��-;8�7^�3/4�0���� ;xZ %&'0B�O� Bä�'#�(c)0�!X�/��G�#--���,�or�� S(3/4�ȤP����� 2(tm)ØFI����"��#���EFZÚ0�۰"��?�� ��F�_6Cx ÎgG"���#|mw��ucL��...�� w0�� ��8;����)�F��E���&� �Xï¿½Ø ï¿½_��wpF�0�{bM�x�>%��o�o���a8�jj� &S�,v�bFl.i��2� �@� i#� �F���V �"��ÅP���ÈHdL$����3�_�_ãq�)'�!1/4��{6~d�?2(c)���d��3/4�b1/2(c)���Ç1/4_��_��-��Z��VA�uкh

  2. Tide Modeling

    First it will be assumed it is cosine. Once that is done it is necessary to calculate how much the graph was moved from its original position. In order to determine it, the difference between the standard cosine graph and the translated one needs to be found. In order to do so two of the same points need to be analyzed for example two crests.

  1. A logistic model

    Graphical plot of the fish population of the hydroelectric project on an interval of 30 years using the logistic function model {19}. The model considers an annual harvest of 8x103 fish. A stable population (4x104) is reached. One can conclude that the population reaches a positive, stable limit in the

  2. Maths IA Type 2 Modelling a Functional Building. The independent variable in ...

    Fig 2. Model of 36m high roof structure at the fa�ade 3 points were plotted on the graphing package, Graph, and a function was given to best fit the points. The function given was , where , meaning that the function given perfectly fits the points.

  1. Ib math HL portfolio parabola investigation

    Graph 2: Value of a=2 Graph 3: Value of a=3 Graph 4: Value of a=4 Graph 5 : Value of a=5 Graph 6 : Value of a=6 a X1 X2 X3 X4 SL = X2 - X1 SR =X4 - X3 D=| SL-SR| 1 1.763932 2.381966 4.618034 6.236068 0.618034 1.618034

  2. Parabola investigation. In this task, we will investigate the patterns in the intersections of ...

    Consider the parabola , the lines and. The four intersections between the parabola , the lines and can be found by again using Graphmatica software. The x-values of these intersections from the left to the right on the x-axis: > x1� 1.149 > x2� 1.382 > x3� 3.618 > x4� 4.351 Calculation of SL and SR: SL

  1. IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream

    Interpretation The reasonableness of the data is acceptable for the most part with minor discrepancy. The data is pretty reasonable until it goes beyond 10 or more hours when the amount of drug in blood reaches below 0 with negative numbers which are impossible.

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    the point of inflection, therefore, the general statement had worked, but with a large percentage error. Also, the general statement is very limited to the flat edges of the trapeziums, where the flat points does not allow for it to ?adapt? to the different curves, causing either overestimations or underestimations of the area.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work