# 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

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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 =

a =

Substituting a = to (4)

1296( ) + 36b = 36

-18b + 36b = 36

b = 2

Therefore: a = , b = 2 , c = 0

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

y = x2 + 2x -----(5)

The given equation is graphed as shown below.

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.

Middle

= 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.

The 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

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 =

Substituted a = to (14):

75( ) + b =1

-1 +b =1

b = 2

We get a = , b=2, c= 0

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

Y=x2 + 2x ------(16)

The above equation is graphed below.

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 = x2 + 2x

17.5= x2 + 2x

x2 + 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 = x2 + 2x

20= x2 + 2x

x2 + 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 = x2 + 2x

22.5= x2 + 2x

x2 + 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 = x2 + 2x

25= x2 + 2x

x2 + 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 = x2 + 2x

27.5= x2 + 2x

x2 + 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 = x2 + 2x

30= x2 + 2x

x2 + 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 = x2 + 2x

32.5= x2 + 2x

x2 + 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 = x2 + 2x

35= x2 + 2x

x2 + 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

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

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!!

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

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