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Maths portfolio Type- 2 Modeling a function building

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Math’s portfolio

Type- 2

Modeling a function building

Topic question –

According to the condition provided the height of the building should be in between 50% to 75% of the given width 72 meters and that should be in the range of 36 to 54 meters.

So to start off with I will design the structure with the minimum height and width of 36 and 72 meters respectively and then I will use the general parabola equation which is

y= ax2+bx+c.

As shown in the figure the coordinate of the left x intercept corner, the right corner’s x intercept the vertex of the parabola is (0, 0), (72, 0) and (36, 36) respectively. Since we have all these coordinates we will now form three equation by substituting all of them in the above equation y= ax2+bx+c.

That will give us these results:-

Equation – 1

When x =0 then y =0

Thus 0 = a x 02 + b x 0 + c

Therefore c = 0

Equation - 2

When x = 72 then y = 0 thus

0=5 184a+72b

So a= -72b/5184

Equation – 3

When x = 36 then y = 36

1296a + 36b = 36

Since we know that a= -72b/5184 now i will replace b with this term and find out the values of ‘a’ and ‘b’. Now the equation will look like 1296(-72b/5184) + 36b=36.Thus after using the calculator I get that

b = 2, a = -0.03 and c = 0

y =image35.gif.



Now I am going to find the dimensions of the cuboid with maximum volume which would fit inside this roof structure.

...read more.


(X of B)


(36-x)2 + image10.png


= image12.png


=  image12.png

[(36-x)2- 2592+72x]

= image12.png


h =image11.png

[1296 - x2]

Volume=width × height × length

=2x ×[image11.png

(1296 - x2)] ×150

= 150[image14.png

(1296x – x3)]

Now in order to get the value of V we must first differentiate 150[image14.png

(1296x – x3)] and equate it to 0.

So the differentiation of 150[image14.png

(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 =image11.png

[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.

...read more.


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 = image33.png

x2 + 2x .which is equivalent to 2.5= image33.png

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= image33.png

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.

...read more.

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

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