xbookgirlx Posted January 25, 2011 Report Share Posted January 25, 2011 YEah... youve got to form a general equation for the height of the cuboid ... when the structure of the roof changes ... take all heights between 36-54 and use excel or even your calculator.. But using excel is reallly helpful and fast and also covers use of technology!!Yea but how can i relate the two I treated the roof of the structure as a parabola and then by using the height of the structure i found the equation for the parabola.I used that equation to find the height of the cuboid, but if the height of the parabola changes then the height of the cuboid changes which makes sense but there would have to me an entirely different varible.do i just leave that in or what????sorry im just kind of confused. Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 25, 2011 Report Share Posted January 25, 2011 Ok.. so now you would have found the equation for height 36 .. now assume instead of 36 a height 'h' ( or any other variable) .. now find an equation in terms of 'h' and integrate the equation to find the area of the curved roof structure!! Reply Link to post Share on other sites More sharing options...
Zananok Posted January 25, 2011 Report Share Posted January 25, 2011 I need help urgently! I just noticed my general equation for each height (point 4) is wrong!I need help fast This is what i had before:-(1/h) x^2+(72/h) x Reply Link to post Share on other sites More sharing options...
chickfilla Posted January 25, 2011 Report Share Posted January 25, 2011 Bleh. So glad I'm done with this. Got 15/20. Reply Link to post Share on other sites More sharing options...
sid1729 Posted January 25, 2011 Report Share Posted January 25, 2011 well..... mine was ....→y=H((-x^2)/1296+1) .. . and on solving the i got an equation for H= x 'h' .. .where H is the height of the cuboid..... x is a constant !! and h is the height of the curved roof structure .. which will be values between 36-54 Reply Link to post Share on other sites More sharing options...
chrypton Posted January 25, 2011 Report Share Posted January 25, 2011 Just handed this portfolio to my teacher today. Man! Am I glad I'm done with this. Reply Link to post Share on other sites More sharing options...
Zananok Posted January 25, 2011 Report Share Posted January 25, 2011 Thanks sid1729, i got the result i wanted few minutes after i made that post, but thanks, really appreciated the reply ^^.Regards,Zananok Reply Link to post Share on other sites More sharing options...
Worked Posted January 25, 2011 Report Share Posted January 25, 2011 i can help if ur willing to help me figure out my patterns of lines portfolio ..... inbox me if your willing, and i'll help you out on the cuboid thing....are you willing to help me with my Modeling Portfolio? Let me know Please if you are. Thanks Reply Link to post Share on other sites More sharing options...
pepsiIBsurvivalmode Posted January 27, 2011 Report Share Posted January 27, 2011 Can anyone help me as to HOW I begin this? I really have no idea what to do because I've always been rather bad at Math IAs. Please help, I can help you back with whatever you need! First of all, I need to find out how to get the equation for the shortest and tallest building: get the 1)curve, 2) volume of building, 3)max cuboid volume, and 4) floor space. I'm sure I'm supposed to use integration/derivatives in this somehow....I would appreciate some help SO MUCH!! Reply Link to post Share on other sites More sharing options...
Mahuta ♥ Posted January 27, 2011 Report Share Posted January 27, 2011 Hey there, there is already a thread about this in the math forum. I am sure you will find it helpful. Modelling Reply Link to post Share on other sites More sharing options...
xbookgirlx Posted January 30, 2011 Report Share Posted January 30, 2011 Does anyone know why the ratio of the volume of the wasted space to the volume of the office area is always the same? Reply Link to post Share on other sites More sharing options...
waffleman123 Posted January 31, 2011 Report Share Posted January 31, 2011 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).Once you do this for part 3 and realize the pattern, how do you go about finding an equation? like what do i want to equation to represent? Reply Link to post Share on other sites More sharing options...
Andoria Posted February 1, 2011 Report Share Posted February 1, 2011 Does anyone know why the ratio of the volume of the wasted space to the volume of the office area is always the same?Yes, I think You must start by considering the general expression for all possible parabolas with zeros at +/- 36 and maximum heights within the given range (36<maximum height<54), which I suppose you've already found out: y= (kX^2)/36 + k, where k is the maximum height/point on your parabola...So you know that the general expression for the facade area under the curve can be found by evaluating the integral of this general parabolic function. Multiplying the general expression derived for the facade area with the given length of the building will give you an expression for the volume of the entire building in terms of k...Theen here comes the point... you must fix another general equation for the volume of the enclosed cuboid and this one should be in terms of k as well. You know that the length of the optimized cuboid stays the same and so does the width of it; they are known constants. However, the height of the optimized cuboid changes in accordance with the maximum height of your parabola, k. These two are related to each other; if you are done with the previous points of the investigation you must haved notice that the height of the cuboid is always equal to 2/3k. Then you simply substitute this into your equation for the volume of the cuboid: cuboid width x cuboid length x 2/3k... and you there you have a general expression for the volume of all possible optimised cuboids with respect to k.So, the final part is to set up the ratio, with the general expressions instead of specific values: general expression for entire building's volume- general expression for optimised cuboid's volume/ general expression for optimised cuboid's volumeThe "k"'s will eventually cancel out... and you will get an exact value for the ratio.Hope that I've explained in an understandable way... pretty hard like this Reply Link to post Share on other sites More sharing options...
sid1729 Posted February 2, 2011 Report Share Posted February 2, 2011 yes... the volume of wasted space to the volume of office space will be the same! Reply Link to post Share on other sites More sharing options...
xbookgirlx Posted February 2, 2011 Report Share Posted February 2, 2011 Does anyone know why the ratio of the volume of the wasted space to the volume of the office area is always the same?Yes, I think You must start by considering the general expression for all possible parabolas with zeros at +/- 36 and maximum heights within the given range (36<maximum height<54), which I suppose you've already found out: y= (kX^2)/36 + k, where k is the maximum height/point on your parabola...So you know that the general expression for the facade area under the curve can be found by evaluating the integral of this general parabolic function. Multiplying the general expression derived for the facade area with the given length of the building will give you an expression for the volume of the entire building in terms of k...Theen here comes the point... you must fix another general equation for the volume of the enclosed cuboid and this one should be in terms of k as well. You know that the length of the optimized cuboid stays the same and so does the width of it; they are known constants. However, the height of the optimized cuboid changes in accordance with the maximum height of your parabola, k. These two are related to each other; if you are done with the previous points of the investigation you must haved notice that the height of the cuboid is always equal to 2/3k. Then you simply substitute this into your equation for the volume of the cuboid: cuboid width x cuboid length x 2/3k... and you there you have a general expression for the volume of all possible optimised cuboids with respect to k.So, the final part is to set up the ratio, with the general expressions instead of specific values: general expression for entire building's volume- general expression for optimised cuboid's volume/ general expression for optimised cuboid's volumeThe "k"'s will eventually cancel out... and you will get an exact value for the ratio.Hope that I've explained in an understandable way... pretty hard like this Thank you i eventually figured it out i had to turn it in y the end of the day on mondaybut i did what u explained so i am happy that i ended up finding the correct method Reply Link to post Share on other sites More sharing options...
Depriviated Posted February 5, 2011 Report Share Posted February 5, 2011 hey can anyone explain part 5 to me plz it says "determine the max office floor area for different values of hieght within the given specification" Really appreciate it Reply Link to post Share on other sites More sharing options...
wanted Posted February 6, 2011 Report Share Posted February 6, 2011 I'm stuck with the third part, since changing the height of the parabola changes it's width as well, so the original equation of the parabola with a modified height doesn't work.Any ideas? Reply Link to post Share on other sites More sharing options...
frenziedIB Posted February 8, 2011 Report Share Posted February 8, 2011 (edited) hello guys..!!For the 7th point, where it says to maximise the office space even further by not having the block in the shape of a sing cuboid,do we proceed assuming the facade is placed on the longer side of the base, of point 6??(which is 150 m) I was wondering because the facade on the longer side renders greater area and thus the volume of the office block than the original facade (72m).. which means that the new office block would also have greater volume when constructed under the facade on the longer side of the base.. so it would maximise the office space more than the original facade.or does it matter? I was confused because people seem to have built it on the original facade..any suggestions?? Edit: oops,, I just realized i made a calculation mistake.. it turns out to be that they both have the same value of volume.. haha Edited February 8, 2011 by frenziedIB Reply Link to post Share on other sites More sharing options...
The Economist Posted February 9, 2011 Report Share Posted February 9, 2011 hey can anyone explain part 5 to me plz it says "determine the max office floor area for different values of hieght within the given specification" Really appreciate itI just started writing this IA and I believe that at point 5 they want you to do the following:Divide the maximum height over the minimum (given) height of each room in order to find the number of floors in you building, then find the area for one floor and multiply it with the number of floors found. There it is. Correct me if I'm wrong. Reply Link to post Share on other sites More sharing options...
the.dibster Posted February 11, 2011 Report Share Posted February 11, 2011 I'm a bit confused...I figured out the equation of the parabola when the height is 36...Using algebra and trig, I found that the volume for the cuboid should be 194400 or close if you dont round.But I dont know which derivatives to take and equate to zero...How do I maximize a rectangle inside a parabola? After that, all I have to do is multiply by 150... but how do I get the maximum.Do I do this? Area = 2xy ... Y = -1/36 x^2 + 36 .... and do substitution?If I combine and then take the derivative and equate to zero... I dont get the right answer... where am I going wrong?How did you find the cuboid volume if you don't know the maximum area cuboid (i.e. the cross-sectional rectangle)?Anyways, for the second bullet, I'm assuming you put the parabola shifted up 36 and thus split by the y-axis, with the x-axis representing the ground. Put a point (x,y) on the parabola representing one vertex of the maximum area rectangle. Now, that means the width is 2x, right? (-x to 0 plus 0 to x distance) So y= f(x), and so the volume is 2x*f(x). Now, use calculus to find the maximum value for that equation, representing area, by deriving and setting equal to zero. Now you have x, so double that for the width and plug it in to the original function for your y, then multiply by 150 to get the maximum possible area.Hope that helps!----Although I myself have a question. The last two bullets ask for how to maximize space when not limited to a cuboid (i.e. minimum floor height of 2.5m, with floors near the bottom larger). How exactly do I go about finding out what height and how many of each floor (assuming, I'm not sure if a valid assumption though- whatever way is easier, that all floors are the same height within the building (all 2.5m or 3m and not individually varying) because I have a sneaking suspicion more area can be squeezed out of the model if I don't use 2.5m, and use 3 or so instead)? 'Cause I'm lost Reply Link to post Share on other sites More sharing options...
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