Matematica para modelar un edificio

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We find ourselves having to design a building with a rectangular base 150m long and 72m wide.

For us to model the roof with the given specifications, all we have to do is draw graph in the form f(x)=ax2 + c and input the data. In this case the data given is the height (36m) and width (72m) since the length is always constant (150m) all working out will be shown in a twodimensional way.

f(x)= ax2+36
If the width is 72m, then both x-intercepts should be (-36,0) and (36,0).
f(±36)=0 a(362)+(36)=0
1296a+36=0
1296a=-36
a=-

f(x)=

-36<x<36
0<y<36

Though the outside appearance will be of a curve such as the one shown above, the offices will be built into a cuboid.

In order to calculate the biggestpossible volume of this cuboid we first have to define its area. In this case, being half of it on either side of the y-intercept, the area could be defined as 2xy.

gx=2xy
gx=2x-x362+36
gx= -x318+72x

Next, we find the derivative of g(x) in order to find g(x)=0



dgxdx=0=-x26+72
x1,2=±123
Note: The negative value isuseless for the needed purpose.
(only positive results will be used further on)

Substituting x into g(x):
g123=-(123)318+72(123)
=997.66
Now we multiply the area by the length (150):

997.66150=149649.2cm3

As the possible values of x (height) range from 36to 54, we will find the maximum volume of the cuboid as the height varies.
To do this, the height 36 will be replaced by h:
∴ px=ax2+h

p 36=0=a362+h
a=-h1296

px= -h1296x2+h ∴ a is the constant needed for the graph to have x-intercepts (-36,0)(36,0),
a=-1362h for any value of 36≤h≤54

Using optimization:

The area of the cuboid is 2xp(x).
∴kx=2x -h1296x2+h= -h648x3+2hx

Finding the derivative and equating to 0:
dkxdx=-h216x2+2h

dkxdx=0=-h216x2+2h

x1= 123

As shown above, as h varies, x does not, therefore the maximum volume shall be achieved with the maximumh:
Volume:kx=-h1233648+2h123150
=163150h

The prior stated characteristic can be easily shown by the following table:

height (value in meters) |
|Total Area (m2) | Total Volume(m3) |
| | |
36 | 997.661 | 149649.190 |
37 | 1025.374 | 153806.112 |
38 | 1053.087 | 157963.034 |
39 | 1080.800 | 162119.956 |
40 | 1108.513 | 166276.878 |
41 | 1136.225 | 170433.799 |
42 | 1163.938 | 174590.721 |
43 | 1191.651 | 178747.643 |
44 | 1219.364 | 182904.565 |
45 | 1247.077 | 187061.487 |
46 | 1274.789 | 191218.409 |
47 |1302.502 | 195375.331 |
48 | 1330.215 | 199532.253 |
49 | 1357.928 | 203689.175 |
50 | 1385.641 | 207846.097 |
51 | 1413.353 | 212003.019 |
52 | 1441.066 | 216159.941 |
53 | 1468.779 | 220316.863 |
54 | 1496.492 | 224473.785 |

As the offices are to be built into a block, space is wasted, in order to find out if at some height, the most space will be used, a ratio can be calculated.To find out the total volume of the building we just integrate the parabola’s equation between 0 and 36 (or -36 and 0) and multiply it by two.

Total A=2036-h362x2+hdx
=2-h3623x3+hx036
=2-h3623363+36h--h362303+0h
=224h
=48h

Having the total area, by multiplying it by 150, we calculate the volume.
Total volume =48h150
=7200h

The following table shows the variation ofvolume wasted in relation to volume used as the height varies:
height (m) | Total Area | Cuboid Volume | Total Volume | Volume Wasted | Volume Wasted: Volume Used |
| | | | | |
36 | 997.661 | 149649.190 | 259200.0 | 109550.81 | 0.73205 |
37 | 1025.374 | 153806.112 | 266400.0 | 112593.89 | 0.73205 |
38 | 1053.087 | 157963.034 | 273600.0 | 115636.97 | 0.73205 |
39 | 1080.800 |...
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