say you have a box
is it possible shrink it such that its new volume and new surface area are exactly half the original?
you should be able to solve this
>>8048566
Suppose there were such a ratio,
Then, it is trivial to see, all boxes have equal volume.
A contradiction.
QED.
>>8048569
what do you mean?
>>8048586
Rest assured, he has no idea of what he's talking about.
>>8048596
do you?
halp
>>8048566
Depends on the shape of the box. The smaller box would need a more efficient shape (cube-like), so the bigger box would have to not be a cube to begin with. Needs more info:
1) shape of initial box?
2) can we change the shape?
>>8048608
yes
>>8048608
the ratio of the sides doesn't matter because you can multiply each side by a constant independently.
a rectangular box can become a cube and a cube can become a rectangle
Yes it is.
>>8048614
proof?
>>8048615
he doesnt need proofs, god speaks to him
>>8048615
>proof?
I'm not doing your homework for you. Other anon already gave you a bunch of hints.
>>8048625
which anon
everyone itt was memeing
>>8048566
Let A, B, and C be the old dimensions of the box. Let a, b, and c be the new dimensions of the box. Let us say that by shrink you mean that a =< A, b =< B, c =< C.
Volume = ABC -> 0.5ABC = abc
Surface Area = 2AB+2AC+2BC -> AB+AC+BC = 2ab+2ac+2bc
2abc(1/A+1/B+1/C) = 2ab+2ac+2bc
1/A+1/B+1/C = 1/a+1/b+1/c
But because of the definition of shrink stated above, this cannot be true, as 1/a >= 1/A, 1/b >= 1/B, and 1/c >= 1/C
So the answer is that shrinking cannot give us a halved volume and halved surface area at the same time.
>>8048632
>scales three independent variables by the same factor
You have three independent variables and two constraining equations. The problem is underconstrained.
>>8048632
If the big box is more rectangular and the small box more cubic, it can.
>>8048566
Say the original box has dimensions [math]a,b,c[/math] and the new box has dimensions [math]x,y,z[/math]. There are infinitely many possible solutions. For some fixed [math]x[/math] then
[eqn]y=\frac{-\sqrt{(a b c-a b x-a c x-b c x)^2-8 a b c x^3}-a b c+a b x+a c x+b c x}{4
x^2}[/eqn]
[eqn]z=\frac{\sqrt{(a b c-a b x-a c x-b c x)^2-8 a b c x^3}-a b c+a b x+a c x+b c x}{4
x^2}[/eqn]
or vice versa.
>>8048644
You can have tips on how to approach it, or you can have an answer, no one's going to grant you the miracle of understanding without effort.
>>8048639
Huh? I never scaled any variables. The assumption that "shrink" means a uniform scaling is too strong and makes the problem way too easy. I simply assumed that by "shrink" OP meant that none of the dimensions of the box can grow. This gives a stronger result.
>You have three independent variables and two constraining equations. The problem is underconstrained.
I don't think you understand a single thing I wrote, since I'm not trying to solve for any variables.
>>8048648
except everyone's answer contradicts everyone else's desu
>>8048640
No, I just proved it can't. The only way you can do it is by making at least one dimension of the box grow longer, and then you aren't shrinking the box are you?
>>8048632
what's happening on the 3rd and fourth lines here?
>>8048655
I simply combined the equation from Volume and the equation from Surface Area. If you really want it spoon fed:
0.5ABC = abc
AB = 2abc/C
AC = 2abc/B
BC = 2abc/A
So AB+AC+BC = 2abc(1/A+1/B+1/C)
2abc(1/A+1/B+1/C) = 2ab+2ac+2bc
Then divide the right side by 2abc and we get the final result.
>>8048652
They don't actually contradict, they just start from different assumptions of what the problem is asking.
>>8048663
your result implies that if
1/A+1/B+1/C=1/a+1/b+1/c then the volume and surface area are half that of the original.
but this is not true since without scaling the volume and area don't change
>>8048708
I don't understand what issue you are having. Any change from A to a can be considered a scaling in one dimension. I proved that any such changes which can be considered "shrinking" cannot lead to both volume and surface area being halved.
Close the lid of the box.
>>8048653
Define smaller faggot. When it comes to boxes, size is determined by volume alone.
>>8048750
read the thread, faggot
>>8048750
>Define smaller
I already did, retard.
>>8048775
So making the box longer is shrinking it? Retard.
>>8048780
So rectangle B isn't smaller than rectangle A?
>>8048569
A box with zero surface area and zero volume works, you silly goose :^)
>>8048791
How does that follow from anything I said? Seriously, explain your "logic".
>>8048800
The width of B is more than A. You say all dimensions must be smaller for it to count as smaller.
I say a smaller box can have one of its dimensions be larger and still be considered the smaller box.
>>8048817
All dimensions of B are smaller than dimensions of box A, retard. My proof does not care about the orientation of the box. Also, the question says that the box *shrinks* not simply that it has a smaller volume. Shrinks implies a continuous scaling that does not allow growth in any dimension.
>>8048831
prove it
>>8048829
Box B has a larger width....
I could have also made B into a square with a slightly larger side than the width of A. It would still look smaller. That way no dimension of B is smaller than A's width.
Most would not make the assumption that you have made. That's why people were asking if you could change the shape.
>>8048838
>Box B has a larger width....
Wow you really are this dense... Flip box B around and both the width and the length are smaller. The orientation of the box has no bearing on its relative size, and nothing in my proof has to do with the orientation of the box. 'A' and 'a' can be any two dimensions as long as they are distinct from the others.
>I could have also made B into a square with a slightly larger side than the width of A.
Then you are still proving nothing as only an idiot would say that you could shrink A into B. All you have done is ignore the wording in the problem and jabber on about size as if area or volume is the only possible meaning when 'size' is not even relevant to the problem. Now fuck off.
>Most would not make the assumption that you have made.
Most people are not very good at interpreting problems mathematically.
>That's why people were asking if you could change the shape.
You can change the shape, you just can't grow in order to shrink.
>>8048852
Better?
>Most people are not very good at interpreting problems mathematically.
"Shrinking" is not a mathematical concept. Most people define box size by volume. Your opinion doesn't change that.
>>8048838
Let me give you an example of why arguing that "Most would not make the assumption that you have made" is stupid:
You are a biologist traveling in a tropical rainforest when you are bitten by a poisonous snake. Luckily you know of a species of frog whose females secrete the antidote to your bite. Even more luckily, you see one of these frogs in front of you. You also know that the male in this species has a distinctive croak and that the population is split evenly between males and females. You hear a croak from a male frog coming from behind you and you turn around to see two more frogs of the species. You only have enough time to run to the lone frog you first saw or the two frogs behind you and lick them before the poison knocks you out. In which direction do you run and why?
>>8048566
I believe it's possible, but the algebra is horrific. I started with x, y, and z and let dx=-px, where p is fixed between 0 and 1. Solving for dy and dz is ugly as hell, but it can be done I think.
>>8048867
>Better?
See "Then you are still proving nothing as only an idiot would say that you could shrink A into B. All you have done is ignore the wording in the problem and jabber on about size as if area or volume is the only possible meaning when 'size' is not even relevant to the problem. Now fuck off.", in my previous post.
>"Shrinking" is not a mathematical concept.
That's why it has to be interpreted mathematically in order to use it in a math problem, my slow friend.
>Most people define box size by volume.
Completely irrelevant. We already know the volume of the box shrinks as its volume is halved. The question is what is meant by shrink.
>>8048875
>only an idiot would say that you could shrink A into B.
Nice argument.
>That's why it has to be interpreted mathematically in order to use it in a math problem, my slow friend.
And you are the almighty interpreter I guess.
>Completely irrelevant. We already know the volume of the box shrinks as its volume is halved. The question is what is meant by shrink.
Exactly. It's not a mathematical concept. You even said earlier that you had ASSUMED a definition. Let OP clarify the rules.
>>8048868
Do the two frogs seem to be poised aggressively? Male frogs will fight over the affections of a female frog, if they're poised aggressively, it means they're both males and the other is a female, if they're relaxed, it means the other two are probably females but you have a better shot of that being the case with the two frogs, especially if the croaking frog seems aggressive but the one beside him isn't that would imply the other frog is a potential aggressor and he's defending his mating rights. Alternatives if they could all be males and you're straight fucked.
>>8048566
The question is immaterial. Regardless of the size of the box, as I am a variety of cat, I will attempt to climb into the box. This operation may be effected by means of the diagram
[math]
\displaystyle
\begin{array}{ccc}
\mathscr{L} & \rightarrow & \mathscr{L}' \\
\downarrow & \searrow & \downarrow \\
\mathscr{L}'' & \rightarrow & \mathbb{B}_{ \mathscr{L}} \\
\end{array}
[/math]
where the script L's denote initial and possible intermediary positions that I may assume with respect to the box, while the terminus B_L denotes the state of affairs once I have actually successfully climbed into the box.
Which will happen, no matter what, least of all with regard to the size of the box. After all, I am a cat.
>>8048891
I will not give you any more information than what was already stated. Do not assume more than was given. This is one reason why the common assumption =/= reasonable assumption. One of the common responses to this problem is to make up behavior of male and female frogs, which is rather stupid. It's an imaginary species.
>>8048890
>Nice argument.
Says the guy who just tried to argue that the common assumption = reasonable assumption.
>And you are the almighty interpreter I guess.
Have some self-awareness. You are trying to assert your interpretation of the problem as much as I am. The only difference is that you are interpreting words that are not even in the problem.
>Exactly. It's not a mathematical concept. You even said earlier that you had ASSUMED a definition. Let OP clarify the rules.
Oh, so you were letting OP clarify the rules by asserting that shrinking only applied to volume? OP is not around and a reasonable assumption is required to answer the question. We were arguing over what is the reasonable assumption and now you are attempting to backtrack to whether an assumption should be made at all because you don't like the result. Have some backbone.
>>8048902
>Underated post
>>8048928
You should read the question again. Maybe you need to read it several times.
>>8048569
"QED"
kek
if by 'shrink' you mean the dimensions stay congruent then no, because the volume is always decreasing faster than the surface area.
but if you're allowed to change the shape then it would work...
if you have a 1x1x2 box and split it to 1x1x1 the volume goes from 2 to 1 but the surface area only goes from 8 to 6, so for it to work you'd have to stretch out the original box to have a surface area of 12
12 = 2x + 4/x
5.something, I fagot how to do that problem
i guess this means the anon that wrote the >>8048632 was right?
>>8049759
No.
>>8048566
no, unless you consider a construct of which 2 sides have 0 surface area a box
>>8049793
Yes, unless you think that "shrinking" means "growing".
>>8049812
>considering the degenerate case
>>8049822
then no
1/2 is 1/2
so yes
i know what you're trying to say but you said it correctly, hence i have the correct answer to the way you presented it instead of how you meant to
try again
>>8049255
>because the volume is always decreasing faster than the surface area.
Nonsense
>if you have a 1x1x2 box and split it to 1x1x1 the volume goes from 2 to 1 but the surface area only goes from 8 to 6, so for it to work you'd have to stretch out the original box to have a surface area of 12
More nonsense
>>8049833
It's wasn't the best use of terminology, but their point was still entirely comprehensible.
>>8048903
male male
male female
female male
If you had to choose one of the two frogs, you would pick neither since there is a 2/3 chance you would pick a male
Since you get to lick both, there is a 2/3 chance you pick a female
So, you pick the two behind you
This assumes that the failure to croak of the first frog has no bearing on the chances of it being male or female
>>8049821
>Yes, unless you think that "shrinking" means "growing".
Under your autistic definition, "growing" means decreasing by a factor of 2.
>>8049867
Ah and here we have the second proof that the common assumption is not the most reasonable assumption. A very common assumption made by people trying to answer this question is that all the possible events are equally likely. This ignores the fact that the lack of a croak is almost as informative as a croak. Interestingly, you recognized this possibility but simply chose to assume it has no bearing. But this assumption makes no sense because in order for the lack of a croak to not effect the chances of being male or female, a female would have to be just as likely to give the distinctive male croak as a male. But we already know that the croak is distinct to males and that therefore females never make this croak. So it's a direct consequence of the problem that the lack of a croak tells us the frog is more likely to be female than male.
Specifically, if a male frog had x chance of croaking while we were listening, then the chance the lone frog in front of you has the antidote is
(1/2)/(1/2+(1/2)(1-x)) = 1/(2-x)
And surprisingly enough, the chance that one of the frogs behind you has the antidote is
2(1/2)(x)(1/2)/(2(1/2)(x)(1/2)+2(1/2)(x)(1/2)(1-x)) = 1/(2-x)
So it doesn't matter in which direction you run. This is because, unlike in the Girl Boy Problem, the male frog croaking is information about a specific frog being male, and thus the case with two frogs reduces to the case of the lone frog which did not croak.
>>8049892
>Under your autistic definition, "growing" means decreasing by a factor of 2.
The box being lengthened by some factor is not a decrease. You're just ignoring the context of what I'm saying and substituting your own in order to come up with a nonsensical strawman.
"I grew by 2 inches this year"
"You couldn't have grown, you shrank by 10 pounds"
What is the question, again?
V = 4/3 π R3
v = 4/3 π r3
v = (8/27)V
Now use this last equation to tie together the two volume equations:
4/3 π r3 = (8/27) (4/3 π R3)
so when you simplify and take the cube root,
r = 2/3 R
From your pic of an open box, I can only assume that by volume you mean the volume of the material the box is made from and not the volume of the interior. In this case, the problem is easily solved by using cardboard of infinitesimal thickness. This way you can half the surface area by scaling length by a factor of 1/sqrt(2). The initial volume is 0, so the final volume is halved as well: 1/2*0 = 0. What do I win?
>>8050823
If you're going to assume shit about frog behaviour then it is no longer mathematically meaningful
>the male frog croaking is information about a specific frog being male
no it isn't
>>8050823
By the way, I don't know where or how you came up with that autistic answer but i found this to help you understand
http://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott
>>8051955
Where did I assume anything about frog behavior? The only information I used was what was given in the problem, that male frogs have a distinctive croak. The fact that you did not think about the consequences of what this means fro frogs which were not heard to croak is a lesson on the limits of basic intuition and trusting the most "common" answer to a problem.
>>8051962
How I came up with the answer is basic knowledge in conditional probability. I'm glad you found the video which I took this problem from, because it illustrates my point even more. The video is incorrect as it attempts to replicate the Boy Girl Problem but does not realize that the croaking mechanism it introduces changes the problem considerably. I actually emailed Dr. Abbott about this and he agreed and wants to change the video. I suggest you email him if you want an "authoritative" confirmation of my answer.
>>8048791
No, it's an optical illusion.
Box B is just further in the background, so it only LOOKS smaller.
>>8051955
>>the male frog croaking is information about a specific frog being male
>no it isn't
Stage 1 of accepting you are wrong: DENIAL
Solution: Explanation
Let us first look at the Boy Girl Problem. You are told that Jim has two children, at least one of which is a boy. You construct the following probability space:
1. MM
2. MF
3. FM
Because males and females are equally distributed in the population, you know that each of these events is equally likely. Thus the probability both Jim has a daughter is 2/3.
Now we do the same for the frog riddle. Note that we only heard croaking from one frog. So we are trying to calculate the probability that the pair contains a female given we heard a frog croaking. The probability space looks like this:
1. M(croaked) M(didn't croak)
2. M(croaked) F
3. M(didn't croak) M(croaked)
4. F M(croaked)
We can immediately see a difference with the boy girl problem, not simply that the probability of a male frog croaking is relevant now, but that the event spaces show a different kind of symmetry. There is no analogous case to MM, because here we know one and only one frog croaked, while in the Boy Girl Problem "at least one is a boy" does not refer specifically to any particular child, and allows that both children are boys. This is where the difference in the two problems arise.
So how do we now calculate the chance of the pair containing a female given we heard a frog croak? It's simply the probability of the pair containing a female and one of the frogs croaking divided by the probability of a frog in the pair croaking. Let's refer to the probability of the events listed above by P1 for event #1, etc.
( P2+P4 ) / ( P1+P2+P3+P4)
2(1/2)(x)(1/2)/(2(1/2)(x)(1/2)+2(1/2)(x)(1/2)(1-x))
1/(2-x)
The next stage is BARGAINING. Let's see you try to come up with ways to disprove what I'm saying.
