Now using the equation, y =
(1296-x2) we will find the height of the cuboid.
Therefore y =
(1296-(20.78)2)=24
Height = 24meters
Thus the Volume of the cuboid 150 41.56 24=149616 m3
So this is the proper dimension of the cuboid
Height = 24m , Width=41.56m, Length=150m
This procedure is only used when the height of the curved structure was 36 m but now I will develop another formula in which the height of the structure will be “h”.
Now that the height of the structure is “h” therefore the coordinate of the top corner, right corner and the base will be (36,h),(72,0) and (0,0) respectively.
Therefore sine we know that these coordinates fall on 3 points of a parabola we will have 3 equations.
C=0
5184a + 72b = 0
1296a + 36b = h
Thus :
5184a + 72b = 0
5184a =-72b
a =
a =
thus put (
in place of “a” in 1296a + 36b = h
therefore
1296(
) + 36b = h
-18b + 36b = h
18b = h
b=
Since we know that b=
therefore we will substitute b in a =
This is equal to
Therefore the equation formed for finding the “H” of the structure will be y=
x2 +
x .
Now I will find the dimension of the cuboid with the maximum amount of volume that would fit inside the roof structured.
Therefore our coordinate of A,B,C and D will be=(36-V,0), (36-V, y(36-V) ), (36+V, y(36+V) ) and (36+V,0) respectively. Since B and C fall on the structure therefore “H”= Y (36-V).
Now in order to find the volume of this cuboid I must use this formula (y) =
x2 +
x)
Therefore Y of b = Y(X of B)=
(X of B)2 +
(X of B)
=
(36-x)2 +
(36-x)
=
[(36-x)2-72(36-x)]
=
[(36-x)2- 2592+72x]
=
[1296-72x+x2-2592+72x]
h =
[1296 - x2]
Volume=width × height × length
=2x ×[
(1296 - x2)] ×150
= 150[
(1296x – x3)]
Now in order to get the value of V we must first differentiate 150[
(1296x – x3)] and equate it to 0.
So the differentiation of 150[
(1296x – x3)] = 1296-3x2 = 0
Therefore x =±20.78m
Again the above value represents either maxima or minima thus again we will have to check wether the answer is + or – and to do this I will differentiate- 3x2= 1296. Which is = 20.78m. therefore the cuboids width=2x 20.78 that = 41.56m
As we can see that the width of this cuboid and the one with the height of 36 m is just the same I can conclude that the width does not change with the change in its height
Now using h =
[1296 - x2] we will find the height of the cuboid .
Using my calculator I get an answer of 0.67h.
So it is clear that the length and width have nothing to do with the increase in the volume of the cuboid and the only thing affecting the volume is the height therefore I have prepared a chart which varies from 50 to 75 percent of 72 m which is beginning with 36 and ending with 54 meters.
Now I will calculate the ratio of the volume of the wasted space to the volume of the office block for each height above.
Waste space= structure’s volume (length of the structure x rectangle ABCD’s area) – office block’s volume (lxbxh).
To find the area of the rectangle ABCD we must integrate (y=
x2 +
x) from 72m to 0m
=
2 +
)x dx
= - h (
now replace x with 72
= -h (
-
)
= -h (96 – 144)
= -h(-48) = 48h
Therefore the structures volume = 150 x 48h
=7200h
Now the volume of the office = 0.67h x 150 x 41.56
= 4176.78h
Now I will calculate the ratio of the space wasted with the increase in height starting from 36m to 54m.
This table points out the amount of space wasted in the office block for different scales of height.
Now to determine the total maximum office space floor are in the block for different value of height within the given specifications.
Therefore area of the floor = 41.56 x 150
= 6234 m2
So the maximum floor area = no. of floors x 6234 m2
This table shows the no. of floors that can be made in a building with varying in height and thus from the table it can be seen that area increases with the no. of floors that can be made.
Given that the base remains the same only the façade is placed on the longer side of the base. So now I am going to investigate what happens.
Now in this configuration the width of the rectangular base = 150m
Length of the building = 72m
The structures height will now be between 50 to 75 percent of 150
Therefore we will assume that the coordinate of the left base corner = (0,0)
Top corner = (75,75)
Left bottom corner = (150,0)
Since we know that these points fall in the parabola.We can form 3 equations from this ax2+bx+c equation.
So when x=0 y=0
Therefore C=0
So when X=75 Y= 75
5625a + 75b= 75
When X= 150 Y= 0
22500a + 150b = 0
Which is equal to (5625a + 75b= 75)- (22500a + 150b = 0)
a =
Substituted
in 5625a + 75b= 75 in place of “a”
b = 2
Therefore a =
, b=2 and c= 0
Thus after finding out all the above terms we can derive this formula Y=
x2 + 2x and its graph is given below.
Now I will find an equation in which the height of the structure will be “h”.
Hence the equation of both the corners will be the same. Again we will use ax2+bx+c
Thus 150a + b = 0
Therefore b=-150a
So when the coordinate of the top is (75,h) then the equation = 5625a + 75b = h
Now I will substitute -150a in place of “b” in the above equation
Which will give us 5625a – 11250a = h
Therefore a=
Now again I will use
and substitute in 150a + b = 0
Thus it will give us b =
Therefore the equation formed will be y=
x2 +
x
Now after completing this I will find out the largest possible cuboid’s dimension that can be fitted in the curved structure with a height of 75m.
It should look like this
A= (75-x,0)
D = (75+x,0)
And because B and C are on the parabola there formula will be B= (75-v, y(75-v)) and
C = (75+x, y)
And the cuboids height will be y and I will denote it with “h”
We will use this formula y =
x2 + 2x and this as its value B= (75-x, y) in order to find the cuboids volume.
Thus it will give us y(75-x) =
(75-x) 2 + 2(75-x)
=
(5625 - x2)
Therefore the volume of the cuboid = 72× 2x ×
(5625 - x2) which can be deduced to
( 5625x - x3) and in order to get the cuboids maximum volume we must differentiate with regards to v
( 5625x - x3) this formula and equate it with 0.
(5625x -3x2) = 0
So x = ±43.3
Again we are unsure whether the value is a maxima or minima thus we can differentiate
(5625x -x3) and after differentiating we will get the value of “x” as -6x which is equal to -259.80.
Since the value of x is in negative It proves that the value is maximum. Thus the cuboids width = 2 x x = 86.6 m
Height =
(5625 - x2)
h = 50m
Volume of the cuboid = 72×86.6×50
=311760m3
After this I will now find the maximum volume of the cuboid with height “h” and he coordinate of A,B,C and D are (75-x, 0), ( 75-x, y), (75+x, y))and (75+x, 0) respectively. The height of the cuboid is y(75-x) thus replace (75-x) in x’s place in the formule y=
[
x2 + 2x] .
=
((75-x) 2 -150(75 – x))=
(5625 - x2)
Therefore the volume =
(5625 - x2) × 2x × 72
Thus to get the cuboids maximum volume we must differentiate this equation
(5625 - x2) × 2x × 72 and then equate it with 0.
After differentiating we will get 5625-3x2 =0 as an answer. Therefore making x = ±43.3m
Therefore the width is 2 x x= 86.6
As we can notice that the value of the volume is the same as above hence it can be concluded that the width does not change with the change in height.
Now in order to find h we will use this formula
5625 - x2
And that is equal to 0.67h.
Thus it’s the height of the cuboid that affect the volume of the cuboid not the width.
the table shown underneath determines the height and the volume of the cuboid for various heights which starting from 50 to 75 percent of 150.
From the table it is clear that the cuboids height increases with the increase of the structure’s height and it also affects the volume of the cuboid.
Now iam going to calculate the amount of space wasted and compare it with the volume of the office with different height. Therefore Volume of wasted space= volume of the structure – volume of the office block..
Office blocks volume = LxBxH
= 86.6× 0.67h × 72 = 4177.58h
Structures volume = length of the structure x its area and in order to find the area we must integrate
[
x2 + 2x] from 0 to 150.
=
=
= Area = 100H
Therefore the volume of the structure = 72 x 100H and from this I will find the amount of space wasted as the height varies from 75m to 112.5 m.
This table shows us the amount of space wasted for different height of the structure. But now I will calculate the total office area with difference in height.
Given that the minimum height of the floor is 2.5m
H= 0.67h
Therefore the area of the floor = 72 × 86.6=6235m2
The total maximum area of the floor = no. of floors x 6235m2
Now by using the above information I can find the total maximum area of the floor and the maximum numbers of floors that can be made in the building with varying height.
This table points out the number of floors that can be made in the building with height as an increasing variable and trough this table it is clear that the maximum office area is directly proportional to the number of floors that can be built.
In this table the 14 number of floors could be made at the max if the building has a height of 36m and when the height of the structure is 75m the maximum number of floors that we can fit is 30.
Now if we concentrate on the design we will see that the floors are made in layers which mean that the base of the structure will be the having the longest base and as we move further on top the floors base would decrease.
Now I will check the maximum amount of space that can be covered in a structure with the height of 36m and length of 150m.
As the cuboids height is fixed as 2.5m now we will find the width and the volume of the lowest cuboid in the structure by using y =
x2 + 2x .which is equivalent to 2.5=
x2 + 2x.Thus after solving this quadratic it we will get x as 1.27, 70.73.
Therefore the base cuboids width = 70.73-1.27=69.46m
And its volume = 150×69.46×2.5=26046m3
As we advance upwards we will have to keep adding the height of the cuboids 2.5m until we reach the total height of 35 as at that time we will have 14 floors
Thus when y = 35 the equation formed will be similar to the one used in the above example thus it will look like this 35=
x2 + 2x and it will be equivalent to x2-72x+1260=0
Thus after solving this quadtric equation we will get x as 30 and 42.
Therefore the width of the base = 42m-30m = 12m and the volume will be equal to 150×12×2.5=4500m3
Thus the Total volume of all the cuboids calculated was =246386m3
For the height of 36m the volume of the structure is= 7200h
= 7200×36=259200m3
Therefore the amount of space Wasted = 259200-246386=12814m3
this is the ratio of the space wasted with the total space
Thus I would conclude by saying that the amount of space wasted with having just a single cubeoid was 0.72.Whereas if we used the above technique we will save more space compared to the previous technique thus it is ethical and economical.
by
avinash166 (student)
