Boom! Part ii will be attempted tomorrow...(Original post by TheMagicMan)
Spoiler:ShowThis is correct. Anything beneath 2 can be reached, but 2 itself cannot.
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 3841
 29032012 01:11

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 3842
 29032012 01:28
(Original post by Slumpy)
After a little bit of work, I'm guessing atSpoiler:Showi2
Anything up to 2 can be won(1/3,0,1/3,1/3, then goes to 0,0,1/3,0 goes to 2/3,0,2/3,0 then repeatedly split the 1 you add between the two places so they're equal). Once above x1, add 1 to the number left. I'm not sure if you can reach 2 however, pretty sure you could make some argument about any number above 1 should be sent to 0 by B, and you can't get more than 1 there at once. Close?Last edited by Blutooth; 29032012 at 01:30. 
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 3843
 29032012 09:18
(Original post by Blutooth)
I have a simple proof that the value must be 2 or below, only a couple of line for ii. 
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 3844
 29032012 22:21
(Original post by TheMagicMan)
..
Looking at that, considering the optimality of B and the number of quantities, I reckon it is fairly easy to build a geometric progression?
You can easily build a symmetry so that B doesn't have a choice, and then every consecutive turn A does not have a choice.
So, if you start with , next you have ... hence, general term and the limit of that is 1.
Therefore, taking into account the final turn, all values below 2 can be obtained.
For part two, I'm getting some strange series. Would you reckon you get a lower value for than 2?
If so, I can try to find something about the third part, but I have to finish few pages of a coursework before that tonight. And to EAT  always forget that.Last edited by gff; 29032012 at 22:23. 
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 3845
 29032012 23:01
(Original post by gff)
Have to finish couple of things these days, so I'm half missing.
Looking at that, considering the optimality of B and the number of quantities, I reckon it is fairly easy to build a geometric progression?
You can easily build a symmetry so that B doesn't have a choice, and then every consecutive turn A does not have a choice.
So, if you start with , next you have ... hence, general term and the limit of that is 1.
Therefore, taking into account the final turn, all values below 2 can be obtained.
For part two, I'm getting some strange series. Would you reckon you get a lower value for than 2?
If so, I can try to find something about the third part, but I have to finish few pages of a coursework before that tonight. And to EAT  always forget that. 
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 3846
 30032012 21:26
(Original post by TheMagicMan)
On the 5 quantity one the maximum value is at least what it is on the 4 quantity one as you can play the same strategy ignoring one quantity if you want
for an even number n, i think the highest value you can get for x is 1+ 1/2 +1/3 +1/4+....+1/(n/2) +1/2
or if the number is odd 1+1/2+1/3+...+1/((n1)/2) +1/2
Can't prove this is the highest though, though looking at the method I can sort of see why this number should be the highest, ie at each step you are focusing on filling fewerpots and at the same time minimising the amount of water thrown out by the thrower until you have one pot remaining.Last edited by Blutooth; 30032012 at 21:37. 
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 3847
 03042012 00:33
Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0?
(Original post by raheem94)
... 
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 3848
 03042012 00:49
(Original post by KingPanther)
Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0? 
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 3849
 03042012 00:49
(Original post by KingPanther)
Hello, what is meant differentiate with respect to x, does that mean calculate when dy/dx = 0?
If you are said to differentiate the above equation with respect to 'x', then you need to find dy/dx.
You still look to be confused about that turning points concept.
When you are asked to find the turning points, then you need to first find the dy/dx (i.e. differentiate the 'y' expression with respect to 'x') and then make dy/dx equal to zero. 
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 3850
 03042012 00:52
(Original post by raheem94)
If you are said to differentiate the above equation with respect to 'x', then you need to find dy/dx.
You still look to be confused about that turning points concept.
When you are asked to find the turning points, then you need to first find the dy/dx (i.e. differentiate the 'y' expression with respect to 'x') and then make dy/dx equal to zero.(Original post by gff)
..
The equation is y=x^1/2(8+x)Last edited by KingPanther; 03042012 at 00:55. 
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 3851
 03042012 00:55
(Original post by KingPanther)
O.k, so I differentiate to find x? 
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 3852
 03042012 00:55
(Original post by raheem94)
Can you tell me which question are you talking about? 
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 3853
 03042012 00:57
(Original post by KingPanther)
The equation is y=x^1/2(8+x) 
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 3854
 03042012 01:01
(Original post by raheem94)
This is the equation, what does the question asks you to find? What is the complete question?
i got, dy/dx=(4x^1/2)+(3/2x^1/2) 
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 3855
 03042012 01:05

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 3856
 03042012 01:06
(Original post by raheem94)
Yes, you did it correctly and got the correct answer . 
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 3857
 03042012 01:09
(Original post by raheem94)
Yes, you did it correctly and got the correct answer . 
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 3858
 03042012 01:13
(Original post by KingPanther)
What would I sub in to get 0? Or do I not have to do that?? 
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 3859
 03042012 01:16
(Original post by raheem94)
I have seen the microsoft word file. You question is done, you don't need to do anything further.
So when gradient is 13, I did
13 = 12x + 5
which is x = 2/3
I dont understand Q)15.Last edited by KingPanther; 03042012 at 01:20. 
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 3860
 03042012 01:20
(Original post by KingPanther)
The next question, I got dy/dx = 12x + 5
So when gradient is 13, I did
13 = 12x + 5
which is x = 2/3
This is correct as well, i also get the same answer.
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Updated: August 3, 2015
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