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Math Problem


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#1 Ernestt

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Posted 13 February 2017 - 11:20 AM

Hi Appreciate if someone can help to offer a solution according to the question shown on the photo? Thks.

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#2 Francisco Angel

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Posted 13 February 2017 - 12:50 PM

Dear Ernestt:

Some context would had been helpful. It not seems to me a force/tension balance problem (?). If that's not the case, I would go like this:

Lets call x the angle between the diagonal chain and the horizontal. The vertical section of the chain is:

 

vs = 2 - 2 *tan(x) = 2 (1- tan(x))

 

The diagonal section of the chain is:

 

hs = 2 / cos(x)

 

The total chain length is (there are two diagonal sections):

 

ts = 4 / cos(x) + 2 (1- tan(x))

 

You can now determine the optimal value of x that minimizes the total lenght ( I leave this to you). Also consider the range of values that x can have. See the attached image.

https://drive.google...mdmRlMxeU5pWUhn

Best regards.



#3 Ernestt

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Posted 13 February 2017 - 10:10 PM

Thank you Mr Francisco for your generousity sharing.... ...
Regards,
Ernest

#4 Profe

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Posted 14 February 2017 - 08:35 AM

Hi Ernestt

 

If you remember the postulate of elementary geometry: The shortest distance between two points is the line that joins these points.
Then: ACD> AD and the shortest length will be AD + DB

 

According your diagram the shortest length is  = 4 * sqrt (2) = 5,656

 

Fausto


Edited by Profe, 14 February 2017 - 08:36 AM.


#5 Ernestt

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Posted 14 February 2017 - 08:40 AM

Hi MrProfe,
Thanks a lot! This helps!
Regards,
Ernest

#6 Francisco Angel

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Posted 14 February 2017 - 12:33 PM

Mr Profe,

 

you are right in that on a "one to one" basis,  the shortest distance will be to avoid the intermediate point C and instead go directly from A to D.  But here, if you take this option, you will double the amount of chain required, because you also must add the segment from B to D.  So there is a possibility for an intermediate optimal solution.

 

Best regards.



#7 Profe

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Posted 14 February 2017 - 02:54 PM

Hi Francisco.

 

As you observe, is twice AD. But the answer already includes the BD segment. That I said: the shortest length will be AD + DB

Good luck

 

Fausto



#8 Francisco Angel

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Posted 15 February 2017 - 08:29 AM


 

Mr Profe:

 

My point is this, using the argument that the straight line is the shortest route, you can say for sure that going directly from A to D will always be shorter than going from A to an intermediate point C and then to D.

 

What you can't say without further proof, is that twice the segment from A to D is shorter that going from A and B to and intermediate point C, and then from C to D, because from C to D you only use a single chain. That is the error of your argurment, in my opinion. If the problem had used two chains in parallel from C to D, your argument would have been valid.

 

Best regards.



#9 Profe

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Posted 15 February 2017 - 12:25 PM

Hi Francisco

 

A postulate is a known truth that does not require proof
Going to the point, the problem does not specify anything for the CD segment, which can vary from 2 units maximum to 0 units in the extreme case.
Therefore it is valid to assume that CD = 0

 

Good luck

 

Fausto

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#10 Francisco Angel

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Posted 15 February 2017 - 06:15 PM

Dear Profe:

 

I think we are not reaching and understanding here. I'm not questioning the postulate you mentioned. I'm not questioning either that it is possible to set CD equal to zero. I'm saying that you can not apply directly your postulate to the problem in question.

Please see the attached Image:

https://drive.google...VpBVEhZNEpNb2dn

Had the problem been like this, with two chains in parallel going from C to D, I would had accepted your solution as optimal without further analysis, please compare this with the image provided by Ernestt and let me know what you think, taking also in consideration my previous posts.

Best regards.



#11 Profe

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Posted 22 February 2017 - 11:51 AM

Hi Francisco

 

You are OK.

 

According to the graph:

If  Chain length = y = 2 AC + CD

CD = ED – EC

AC = SQRT(AE2 + EC2)

If             EC = x, AE = a, ED = b

Then :  y = 2*SQRT(a2 + x2)+(b-x)

Search  y minimum, with a= 2 and b = 2

Resolving we have:

Value of x = 1,15454 thath minimizes y

Value of y = 5,46410 minimum

 

Nice problem to remember Maths!

 

Fausto



#12 Francisco Angel

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Posted 22 February 2017 - 03:20 PM

Profe:

 

I get the same value for the objective function that I wrote in the first post. x=0.5236, minimum length=5.4641. I didn't want to provide the optimal values that time to let some work for Ernestt to do. Nice to see we arrived at the same result.

Best regards.






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