No Rest for the Wicked [Bleach]
- Thread starter Tekomandor
- Start date
Problem with this bit of math. 100 mm * 100 mm =/= 10 mm2
The conversion rate between square and cubic meters is not the exact same as their normal varient. There are 1000 mm in a meter, but 1000000 mm2 in a meter2
So that's at least 60,000 mm2 for all 6 sides.
And each flap (100 mm long and 5 mm wide) are 500 mm2.
Last edited:
We would presumably be cutting it to the right shape. But that makes me wonder, is it asking for the area of the resultant storage cube, or of the rectangular piece of cardboard it's being cut from?
@Tekomandor ?
Edit: It would be 63,500 mm2 if we are just talking about the cube, but if we are talking about the rectangle it's cut out from it would be 125,550 mm2
Last edited:
Going by your assumption that we have a scissor, I'm getting the same area for the cardboard that goes into making the cube but a different one for the rectangle it's made of.
EDIT: Argh, wait not even that. Problem is if we optimize for least cardboard total we get different answers for both numbers compared to optimizing for least cardboard used in the cube.
Last edited:
- Location
- Australia
Well, a bit of a sidenote; the whole question is a bit of a spherical cow in a vacuum problem.
The question failed to accomodate for the thickness of the paper, for instance. You need to accomodate the thickness of the cardboard being used, or it's going to throw your measurements off as the cardboard folds down around it.
A litre is a thousand cubic centimetres, or ten centimetres a side (10*10*10 for a total of 1,000). This therefore gives you an area of one hundred cubic centimetres per side. Change this to millimetres, and you have one hundred millimetres a side (10mm = 1cm), or a total area of ten thousand millimetres per side (100mm*100mm = 10,000mm).
This gives us a total area needed so far of forty thousand square millimetres, or four hundred square centimetres.
Then you need the overlap area of 5mm. This needs to run the whole length of the container, so it's not just an added 5mm- it's 5mm*100mm, or 500mm.
Then you need the top and the bottom. This is simple, since you're literally just adding two more squares to the cube. The length and width are each 100mm, so you're adding another total 20,000mm, bringing us to a total of 60,500mm so far.
Then you need to calculate the area of the flaps that are going to be folded down over each the top and bottom. This is just another one of the flaps we did above, since this is all a perfect cube (a square container). 5mm*100mm = 500mm, and we're doing three for each the top and bottom, for a total of 3,000mm.
All up, you'd need a total of 63,500 square millimetres of cardboard.
But, as I mentioned earlier, that's a spherical cow in a vacuum problem, because it ignores some rather vital things. Those measurements are the measurements necessary for the material inside, but the cardboard must necessarily have thickness, which throws out our whole measurements.
If you assume the cardboard is 1mm thick, for instance, we no longer have a cube that is exactly 100mm a side. Rather, when measured, the cube would be 102mm a side- 100mm, plus 1mm thickness.
All the flaps we made have to have an extra 1mm attached to them, too, to fold over the lip of the cardboard. I'm actually pretty sure you wouldn't need exactly 1mm, you'd need slightly more or less, but either way.
So instead of needing a perfect 10,000mm a side, we're needing 10,404mm a side, or 62,424mm for all six sides of the cube. The flaps also increase in size, becoming (6mmx102mm), or 612mm each. We had 7 of them, so that's up to 4,284mm.
So now all up, we need 66,708 square millimetres of cardboard- a fair cry off from the 63,500 we would have needed earlier.
Of course, all that is dependent on the people who wrote the question taking that into consideration and not making me out into a pedantic ass. It also assumes that my seventh-grade maths is actually correct, and given it's been over a decade since I took it, I could be woefully off.
The question failed to accomodate for the thickness of the paper, for instance. You need to accomodate the thickness of the cardboard being used, or it's going to throw your measurements off as the cardboard folds down around it.
A litre is a thousand cubic centimetres, or ten centimetres a side (10*10*10 for a total of 1,000). This therefore gives you an area of one hundred cubic centimetres per side. Change this to millimetres, and you have one hundred millimetres a side (10mm = 1cm), or a total area of ten thousand millimetres per side (100mm*100mm = 10,000mm).
This gives us a total area needed so far of forty thousand square millimetres, or four hundred square centimetres.
Then you need the overlap area of 5mm. This needs to run the whole length of the container, so it's not just an added 5mm- it's 5mm*100mm, or 500mm.
Then you need the top and the bottom. This is simple, since you're literally just adding two more squares to the cube. The length and width are each 100mm, so you're adding another total 20,000mm, bringing us to a total of 60,500mm so far.
Then you need to calculate the area of the flaps that are going to be folded down over each the top and bottom. This is just another one of the flaps we did above, since this is all a perfect cube (a square container). 5mm*100mm = 500mm, and we're doing three for each the top and bottom, for a total of 3,000mm.
All up, you'd need a total of 63,500 square millimetres of cardboard.
But, as I mentioned earlier, that's a spherical cow in a vacuum problem, because it ignores some rather vital things. Those measurements are the measurements necessary for the material inside, but the cardboard must necessarily have thickness, which throws out our whole measurements.
If you assume the cardboard is 1mm thick, for instance, we no longer have a cube that is exactly 100mm a side. Rather, when measured, the cube would be 102mm a side- 100mm, plus 1mm thickness.
All the flaps we made have to have an extra 1mm attached to them, too, to fold over the lip of the cardboard. I'm actually pretty sure you wouldn't need exactly 1mm, you'd need slightly more or less, but either way.
So instead of needing a perfect 10,000mm a side, we're needing 10,404mm a side, or 62,424mm for all six sides of the cube. The flaps also increase in size, becoming (6mmx102mm), or 612mm each. We had 7 of them, so that's up to 4,284mm.
So now all up, we need 66,708 square millimetres of cardboard- a fair cry off from the 63,500 we would have needed earlier.
Of course, all that is dependent on the people who wrote the question taking that into consideration and not making me out into a pedantic ass. It also assumes that my seventh-grade maths is actually correct, and given it's been over a decade since I took it, I could be woefully off.
Last edited:
*bangs head on desk* Thank you.
NeverDies
Classy Coleoptera ಠ_ರೃ
- Location
- Colorado
I feel like if we wrote that down, the science division would hunt us down and give us a high five.
- Location
- Australia
- Pronouns
- She/Her
And all that is assuming it's asking us to optimize for the surface area of the resultant cube, it's possible that it's asking us to optimize for the area of the rectangular piece of cardboard we are cutting it out of.
@Tekomandor ?
@Tekomandor: So I take it we do get a scissor and they want the area of cardboard that ends up making the cube? Because the original question talked about folding.
Edit: Presuming that we're looking to minimize the original cardboard area and we have a scissor, I'm getting 123,525 mm2. If we don't get a scissor I have no idea. How do you fold a rectangle into a cube?
Edit: Presuming that we're looking to minimize the original cardboard area and we have a scissor, I'm getting 123,525 mm2. If we don't get a scissor I have no idea. How do you fold a rectangle into a cube?
Last edited:
- Location
- Australia
Alright then. The answer should be... hm.@Tempera - no, the writers of the test aren't quite that pedantic. Assume the cardboard has no thickness. But I feel that answer is in character for Kanade so
The minimum area of cardboard needed to make the storage container directly is 63,500 square millimetres, cut into the correct shape. The minimum area of cardboard needed if they handed us a correctly-sized piece of cardboard and told us to cut out the shape we needed would be... 125,550 square millimetres.
Of course, I have absolutely zero problems with Kanade being exactly that pedantic in her response.
Nope, incorrect. If we're optimizing for minimum area in the original cardboard piece we can do better then that, since most of the flaps can be taken from "overhanging" cardboard pieces. It's 123,525 mm2.
EDIT: Nevermind, made a mistake. Your answer is correct.
Last edited:
- Pronouns
- she/her
[x] Be aggressive and get in close!
If they're better than us I don't think playing a defensive game will work.
If they're better than us I don't think playing a defensive game will work.
That's assuming we are optimizing for the resultant cube, and not for the rectangular piece it's being cut out of. Which I'm not really sure how to solve for.
The area of the rectangular piece is (2z+2y+5)*(2y+x+10)
We want that number to be as small as possible while x*y*z=1000000
x,y, and z are all in mm by the way.
Last edited:
And that one is assuming the ideal shape to fold the cube is one rectangle in the middle with one rectangle adjacent at each side, except one side which gets two. I'm not certain that's right either.
As for solving it, it works the same way it does in 1 dimension, except when checking whether it's a minimum you need the determinant of the hessian matrix. I don't think that's really what the question meant though, so I won't bother.
Hmm, now that you mention it, a net with a zig-zag shape would fold into a cube, with only a area of 2 squares by 5 squares, rather than most other formations that get a 3x4.
XXX
....XXX
Kinda like that, Xs marking the squares. Which gives the equation for the cardboard area being (3z+x+y+5)*(x+y+5) which should allow for a much more compact final product.
And I have no idea what a determinant of a hessian matrix is, please enlighten me cause it sounds interesting.
It's used in multi-variable calculus to determine the nature of critical points, basically. The matrix is constructed using the second-order partial derivatives arranged with each variable getting a row and a column, I think. The determinant of a matrix is calculated iteratively and is kind of a pain.
For a function of two variables, the determinant of the hessian would be the difference of the product of the second order partial with respect to x and the second order with respect to y and the square of the second order mixed partial. If the difference is positive, that is, that the product of 'pure' second-orders is larger, then that point on the function is a min/max. If it is negative, that is, the square of the mixed second-order is larger, then that critical point is a saddle point. If the difference is zero, nothing can be determined from the test.
Last edited:
So fxx*fyy-(fxy)2?
And how is it adapted to a three-dimensional equation? fxxx*fyyy*fzzz-(fxyz)3?
Plus, if it's just for finding the nature critical points, how would we account for the fact that, since x*y*z has to =1000000, our minimum might not be at what is otherwise a critical point.
Either you use the equation to eliminate one variable, giving you two free ones, or you use a thingy called a lagrangian multiplier.
Calculating determinants in 3d is painful to explain in text form. I suggest a quick google search for some images or a video.
Last edited:
So bonus perception(or whatever represents pedantic)?@Tempera - no, the writers of the test aren't quite that pedantic. Assume the cardboard has no thickness. But I feel that answer is in character for Kanade so