Now the coordinate of the left bottom corner of the base is (0,0), the coordinate of the right bottom corner is (72,0) and the coordinate of the top corner will be (36,h)
Since the above 3 point lies in the parabola, we will get 3 equations.
C=0 -------(9)
5184a + 72b = 0 -------(10)
1296a + 36b = h ------------(11)
Solving:
5184a + 72b = 0
5184a =-72b
a =
a =
Substituting a = to (4)
1296( ) + 36b = h
-18b + 36b = h
18b = h
b=
Substituting b= to a =
a= - , b=
Therefore: a = , b =
So the equation will be
y=x2 + x --------(12)
This is the equation for the curved roof structure of height “h”
I will find the dimension of the cuboid of maximum volume in this curved roof structure.
Coordinate of A=(36-V,0)
Similarly, coordinate of D=(36+V,0)
As the upper corner of the cuboid B and C are lying on the curved roof structure so
Coordinate of B = (36-V, y(36-V) )
Coordinate of C= (36+V, y(36+V) )
y(36-V) is the height of the cuboid “h”
Substituting (36-V) into equation (12)
Y(36-v)=(36-v)2 + (36-v)
= [(36-v)2-72(36-v)]
= [(36-w)2- 2592+72v]
= [1296-72v+v2-2592+72v]
h =[1296 - v2] ------(13)
Volume=width × height × length
=2v ×[(1296 - v2)] ×150
= 150[(1296v – v3)]
To get “v” for the maximum volume of the cuboid, the above equation should be differentiated with regard to “v” and then equate to zero
=150[(1296v – v3)]=0
Therefore 1296-3v2=0
-3V2=-1296
V=±20.78m
This value can be maxima or minima hence I will differentiate the equation (8) again and check whether the result is positive or negative.
=(-6V)=-124.68
As the above value is negative, hence the value of “v”20.78 is for largest cuboid.
Therefore width of the cuboid = 2v = 41.56 meters
The value is same as in the case of 36 meters height structure hence I can say that the width of the largest cuboid does not change in the structure’s height.
Using the equation (13)to find out the height:
h =[1296 – (20.78)2]
h =0.67h
The above relation is used to find the dimensions and the volume of the cuboid for different heights of the structure.
The length of the cuboid= 150 meters
The width of the cuboid =41.56meters
The width and the length of the cuboid do not change with the change in the structure’s height.
Height of the cuboid is the only variable which affects the volume of the cuboid.
The table below shows the height and the volume of the cuboid for different height of the structure.
Here the height of the structure varies from 50% of 72 to 75% of 72 meters that is from 36 to 54 meters.
The above table shows that the height of cuboid increase with the increase in height of the structure. The volume of the structure of the cuboid increases with the increase in the height of the cuboid hence we can say that the volume of the cuboid increase with the increase in the height of the structure.
Now I calculate the ratio of the volume of the wasted space to the volume of the office block for each height mentioned above.
Volume of the waste space= volume of the structure – volume of the office block
Volume of the office block= height × length × width
=0.67H × 150 × 41.56
= 4176.78H
The volume of the structure= length × area of ABCD
= 150× area of ABCD
In order to find the area of ABCD, I then integrate the equation (12)from 0 to 72m
2 +) dx
2) + x] dx
=-H
=-H [( -)-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.
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.
Now I will develop a general equation for this curved roof structure when the height of the structure is “h”
Now the left and the right bottom corner will remain the same hence we have the same equations as earlier:
150a + b =0 ---------(17)
The coordinate for the top will be (75, h)
5625a + 75b = h ----------(18)
Solving the equation:
b = -150a
Substitute b = -150a to equation (18)
5625a + 75(-150a) = h -----------(19)
5625a – 11250a = h
a =
Substituted a = to (17):
150() + b = 0
b =
so the equation of the roof curve structure will be:
y= x2 + x
= [x2 + 2x] ------(20)
Next I will find out the dimensions of the largest cuboid which can be fitted inside the curved roof structure of height 75 meters
The graph of the cuboid fitted inside the cruved
Coordinates of A = (75-v, 0)
Similarly, coordinate of D =(75+v, 0)
As the upper corner of the cuboid B and C are lying on the curved roof structure so
Coordinate of B= (75-v, y(75-v))
Coordinate of C=(75+v, y(75+v))
Y(75+v) is the height of the cuboid and lei it be “h”
Using equation (16), and coordinate of “B”
Y(75-v) = (75-v) 2 + 2(75-v)
= [(75-v)2-150(75-w)]
= (5625-150v-v2-11250+150v)
= (5625 - v2) -----(21)
Volume of the cuboid = length × width × height
Volume = 72× 2v × (5625 - v2)
= ( 5625v - v3) ----------(22)
To get the maximum volume of the cuboid, the above equation should be differentiate with regard to “v” and then equate to zero.
(5625v - v3) = 0
(5625v -3v2) = 0 -------------(23)
10800 - v2=0
V = ±43.3
This value can be maxima or minima hence I will again differentiate the equation (23) and check for the result whether is positive or negative.
=(-6v)=-259.80
As the above value is negative, hence the value of “v” is maximum.
Therefore width of the cuboid = 2v = 86.6m
Using equation (21)
Y(75-v)=(5625-43.32)
= 50
Hence the height h= 50m
Volume= 72×86.6×50=311760m3
So the dimensions of the cuboid are
Height = 50m , width = 86.6m, length= 72m
Now I will find out the dimension of the cuboid of maximum volume in the curved roof structure of height”h”
Coordinate of A = (75-v, 0)
Coordinate of D = (75+v, 0)
Coordinate of B = ( 75-v, y(75-v))
Coordinate of C= (75+v, y(75+v))
Y(75-v) is the height of the largest cuboid
Using equation (20) and coordinate of B
Y(75-v)=[(75-v) 2 + 2(75-v)]
=((75-v) 2 -150(75 – v))
= (5625-150v +v2 -11250 +150v)
= (v2 -5625)
= (5625 - v2) ------------------(24)
Volume = height × width × length
=(5625 - v2) × 2v × 72
=72[5625v – v3)
To get the maximum volume of the cuboid, the above equation should be differentiate with regard to “v” and then equate to zero.
= 72[5625v – v3) = 0
=5625-3v2 =0
V=±43.3m
Width = 2v = 86.6
This value is the same as in case of 75m height structure hence I can say that the width of the largest cuboid does not change with the change in the structure’s height.
Using equation (24)
H= (5625 - 43.3 2)
=0.67h
Therefore the change of the width of the façade does not affect the relation between the height of the structure and the height of the cuboid of maximum volume.
The above relation is used to find the dimensions and the volume of the cuboid for different height of the structure.
The width and the length of the cuboid do not change with the change in the structure’s height.
The height of the cuboid is the only variable which affect the volume of the cuboid.
The table below shoes the height and the volume of the cuboid for different height of the structure.
Here the height of the structure varies from 50% of 150 to 75% of 150 meters that is from 75 to 112.5 meters.
The above table shows that the height of the cuboid increase with the increase in the height of the structure. The volume of the cuboid increases with the increase in the height of the cuboid hence we can say that the volume of the cuboid increases with the increase in the height of the structure.
Now I going to calculated the ratio of the volume of the waste space to the volume of the office block for each height mentioned above.
Volume of the wasted space= volume of the structure – volume of the office block
Volume of the office block= width × height × length
86.6× 0.67h × 72 = 4177.58h
The volume of the structure= length × the area of the structure
To find the area of the structure, I will integrate the equation (20) from 0 to 150
A=
A=
A=) - 0
A= 100H
Volume of the structure= 72×100H=7200h
Using this, I will calculate the ratio of wasted space to the volume of the office blocks for the structure height ranging from 75m to 112.5m
The above table shows that the ratio of the wasted space to the office block is same for different height of the structure.
I will calculate the total maximum office floor area in the block for different height within the given specifications.
Minimum height of a floor = 2.5m
We know that h = 0.67h
Area of the floor = width × length = 72 × 86.6=6235m2
Total maximum floor area = number of floors × 6235m2
Using the above information I will calculate the total maximum office floor area and the maximum number of floors that can be constructed in the building of different height.
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 in the number of floors
We can see that the maximum number of floors that can be accommodated was 14 when the structure was 36 meters, whereas it is 30 when the height of the structure is 75meters.
Now if I have to maximize the office space then I have to have multiple cuboid instead of single cuboid. The cuboid are to be designed in such a way that the lowest cuboid whose base lies on the x axis has the maximum width and its upper edges are touching the parabola. Similarly the next cuboid whose base lies on the previous cuboid’s upper edge and its upper edges are touching the parabola also, and so on.
I will calculate the maximized office space for the curved roof structure of height 36m and length 150m
As the height of the cuboid are fixed as 2.5m,the y coordinate of the upper edge will be 2.5 for the lowest cuboid. The x coordinate can be found out by suing equation (5) of the curved roof structure.
y = x2 + 2x
2.5= x2 + 2x
x2 + 2x -2.5=0
x2-72x+90=0
x=1.27, 70.73
Width of the cuboids base= 70.73-1.27=69.46m
Volume of the cuboid=150×69.46×2.5=26046m3
The next cuboid’s base will be on the upper edge of the previous cuboid. The height of this cuboid will be 2.5+2.5=5, therefore use the same method as before
y = x2 + 2x
5= x2 + 2x
x2 + 2x -5=0
x2-72x+180=0
x=2.593, 69.406
Width of the cuboids base= 69.406-2.593=66.81m
Volume of the cuboid=150×66.81×2.5=25055m3
The height of this cuboid will be 2.5+2.5+2.5=7.5
y = x2 + 2x
7.5= x2 + 2x
x2 + 2x -7.5=0
x2-72x+270=0
x=3.97, 68.03
Width of the cuboids base= 68.03-3.97=64.06m
Volume of the cuboid=150×64.06×2.5=24022.5m3
The height of this cuboid will be 2.5+2.5+2.5+2.5=10
y = x2 + 2x
10= x2 + 2x
x2 + 2x -10=0
x2-72x+360=0
x=5.41, 66.59
Width of the cuboids base= 66.59-5.41=61.18m
Volume of the cuboid=150×61.18×2.5=22942.5m3
The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5=12.5
y = x2 + 2x
12.5= x2 + 2x
x2 + 2x -12.5=0
x2-72x+450=0
x=6.91, 65.09
Width of the cuboids base= 65.09-6.91=58.18m
Volume of the cuboid=150×58.18×2.5=21817.5m3
The height of this cuboid will be 2.5+2.5+2.5+2.5+2.5+2.5=15
y = x2 + 2x
15= x2 + 2x
x2 + 2x -15=0
x2-72x+540=0
x=8.5, 63.5
Width of the cuboids base= 63.5-8.5=55m
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!!
