IA (HL) Portfolio Type II -- Modelling a Functional Building

[teaseme]

I know the volume of the whole building, but how do i obtain the volume equation of the cuboid? How do I obtain the height of the cuboid?

wait how do you find the volume of the whole building?

[∫ Jorge δx]

I recommend making your facade a parabola, it'll be easier to manipulate height.

Also, when I said volume formula, I meant the the volume for the whole building. I'll give you a hint, it's somewhere along these lines V=2xy, where y is the formula of the parabola that defines your facade (obviously, in terms of x).

[teaseme]

would integration be required in order to find the volume of wasted space?

[Ben92]

I recommend making your facade a parabola, it'll be easier to manipulate height.

Also, when I said volume formula, I meant the the volume for the whole building. I'll give you a hint, it's somewhere along these lines V=2xy, where y is the formula of the parabola that defines your facade (obviously, in terms of x).

That's not the volume of the building. that's the area of a rectangle inscribed under said parabola.

[Morpheus]

would integration be required in order to find the volume of wasted space?

Yes

integrate every example you use for proof

[∫ Jorge δx]

I recommend making your facade a parabola, it'll be easier to manipulate height.

Also, when I said volume formula, I meant the the volume for the whole building. I'll give you a hint, it's somewhere along these lines V=2xy, where y is the formula of the parabola that defines your facade (obviously, in terms of x).

That's not the volume of the building. that's the area of a rectangle inscribed under said parabola.

Oops, it's V=150(72)(integral of y).

However, essentially, V is (150)2xy

My bad

Edited December 20, 2010 by qliphoth93

[IB_soldier]

So....... volume of the cuboid equals (150)(2x)(the parabola equation)? thanks a lot......

I know the volume of the whole building, but how do i obtain the volume equation of the cuboid? How do I obtain the height of the cuboid?

wait how do you find the volume of the whole building?

Using integration, find the area under the curve of your parabola. Then multiply that by 150 and you will get the volume of the whole building as the building is a prism.

[∫ Jorge δx]

Don't forget to use derivatives

And also, if you want to make your life easier at the part related to wasted space, compare the area of the facade, given by a defined integral of your parabola, and the front face of your cuboid, given by 2xy (with the value you obtained from your previous derivative in the place of x).

It uses less keystrokes at any rate

Edited December 21, 2010 by qliphoth93

[Morpheus]

Is anyone trying to use excel?

[waleed]

I am doing this for maths at the moment. I have found that to find the values for the equation, you have a simplified equation and input the value 36 because that is the height of the curve and therefore you can find K (when using Y=-kx^2+36). Use this to find out the x value through differentiation and volia

then you can find out the area and volume from there.

I found using a spreadsheet for point 3 works well and in point 4, I found each different height gave the same ratio of wasted space to volume of the office block. Point 5 I haven't worked out what to do with it yet but I have been suggested using levels? but it seems that is more likely for point 7. Point 6 is easy you just have to do points 1 and 2 again using the different values. Hope this helps

Hello ellie. I found the equation. I found the value of k. Now in order to find the maximum possible volume of the cuboid, I am a bit clueless. I read you talking about differentiating and then finding the volume. You think you can please elaborate?

[Ben92]

I am doing this for maths at the moment. I have found that to find the values for the equation, you have a simplified equation and input the value 36 because that is the height of the curve and therefore you can find K (when using Y=-kx^2+36). Use this to find out the x value through differentiation and volia

then you can find out the area and volume from there.

I found using a spreadsheet for point 3 works well and in point 4, I found each different height gave the same ratio of wasted space to volume of the office block. Point 5 I haven't worked out what to do with it yet but I have been suggested using levels? but it seems that is more likely for point 7. Point 6 is easy you just have to do points 1 and 2 again using the different values. Hope this helps

Hello ellie. I found the equation. I found the value of k. Now in order to find the maximum possible volume of the cuboid, I am a bit clueless. I read you talking about differentiating and then finding the volume. You think you can please elaborate?

What math book are you using? Of you are using H&H go to the calculus-derivative section where they have a discussion of the derivatives of the natural exponent, e.

Basically, your model is of the facade of the building, now you want to find the cuboid with maximum volume that would fit inside the whole building, that means it's the largest rectangle that would fit inside your facade model.Now, if you have a rectangle inscribed under a quadratic function, what would be its area?

Think about it. Once you have found the area of the maximum cuboid you multiply it by the length of the building, in this case 150 m, to find the volume of the cuboid.

Then, for the volume of the building itself, you find the area of the facade, i.e. the area under the parabola of the form y=kx^2+36. And once again you multiply that by 150m, getting the volume.

Hope this helps, good luck

http://www.google.co.uk/imgres?imgurl=http://mysite.verizon.net/bnapholtz/Math/images/maxmin2.gif&imgrefurl=http://mysite.verizon.net/bnapholtz/Math/extrema.html&usg=__nqQU18vDVsi4y9C8ERD0iN6ZZA8=&h=182&w=165&sz=2&hl=en&start=61&zoom=1&tbnid=MdxcGOwq63ZrmM:&tbnh=145&tbnw=132&prev=/images%3Fq%3D2xy%2Bparabola%2Brectangle%26um%3D1%26hl%3Den%26biw%3D1280%26bih%3D685%26tbs%3Disch:10,1711&um=1&itbs=1&iact=hc&vpx=1087&vpy=338&dur=88&hovh=145&hovw=132&tx=78&ty=121&ei=9TQWTeufKsfDswbYv-jjDA&oei=7DQWTaK8NYG38gPg96iDBw&esq=4&page=4&ndsp=15&ved=1t:429,r:9,s:61&biw=1280&bih=685

I think this picture helps a lot, if you still haven't understood.

[AAMomen]

Am I supposed to keep getting the same x-values (and therefore, the same width) for each different height I try? It makes sense in my head but at the same time it doesn't really. Help please?

[Ben92]

Am I supposed to keep getting the same x-values (and therefore, the same width) for each different height I try? It makes sense in my head but at the same time it doesn't really. Help please?

yes,you are... don't worry... remember your actual width is 2 times that.

Relax, this is a very easy portfolio in terms of the math just write it well and clearly and show your working and you'll do great

[sweet777]

has anybody tried to use the ellipse equation for this ?

and can anyone confirm that the maximum volume is at x=20.7 ( width = 41.4)

And how do you usd technology to investigate how changes to the height of the structure affect the dimensions of the largest cuboid ??

hey

umm.. I've got a different value. and now I'm scared that I'm probably wrong. my width is more than that. one more question. the second last question, can someone give me a hint on how to go about that?? only a teeny tiny hint?

[∫ Jorge δx]

For the second to last part, I suggest you make the inner structure that looks like the following...

In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine.

However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.

Edited December 26, 2010 by qliphoth93

[Ben92]

For the second to last part, I suggest you make the inner structure that looks like the following...

In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine.

However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.

why is it only 14 (by my count floors)? I mean is that your final model? I got mine to be 17 floors, maximizing floor space.

[why?]

hi guys

I just started the task and I got stuck with the 3 point. What kind of technology should I use?

[ppp]

Am I supposed to keep getting the same x-values (and therefore, the same width) for each different height I try? It makes sense in my head but at the same time it doesn't really. Help please?

same problem does anybody know more?

[AAMomen]

For the second to last part, I suggest you make the inner structure that looks like the following...

In case you're wondering, you cannot use this image for your IA, since I took it straight out of mine.

However, you can make one like that yourself, using Google SketchUp or any other 3D modeling software.

why is it only 14 (by my count floors)? I mean is that your final model? I got mine to be 17 floors, maximizing floor space.

Hey. What height are we supposed to use? The parameters I got were between 36 and 54. So I imagined that to maximize floor space one would use the maximum height, dividing it by the minimum height of a room. This causes me to have about 21 floors. Why is my building so much bigger than your guys' ones!? And I'm zetta proud of it. But really. How am I wrong?

Edit : Oh and I got 14 when I used 36 as the height of the curved structure

Edited December 27, 2010 by AAMomen

[foreverhunched]

Am I supposed to keep getting the same x-values (and therefore, the same width) for each different height I try? It makes sense in my head but at the same time it doesn't really. Help please?

same problem does anybody know more?

This is what I think...

You're only changing height of the structure (width of the structure remains the same). So it makes sense that the width of the cuboid will also stay the same. Hope that helps