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pex
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Minimization problem  
« on: Jan 14th, 2015, 2:27am »
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Minimize the expression
 
(x2 - 39x + 507)1/2 + (y2 - 138y + 6348)1/2 + (x2 - xy + y2)1/2
 
where x and y are real numbers.
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rmsgrey
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Re: Minimization problem  
« Reply #1 on: Jan 14th, 2015, 9:05am »
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x, y arbitrarily large; take the negative roots.
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Re: Minimization problem  
« Reply #2 on: Jan 14th, 2015, 10:35am »
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x = 598/35
y = 299/12
F(x,y) = 93
 
via D(F,x)=0 and D(F,y)=0
Also via genetic algorithm Wink

Still not sure how to do it the easy way.
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Re: Minimization problem  
« Reply #3 on: Jan 14th, 2015, 3:40pm »
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The easy way sometimes the more difficult it is.
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Re: Minimization problem  
« Reply #4 on: Jan 15th, 2015, 8:53am »
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towr's result is correct, obviously. Grimbal's remark doesn't really apply here, which is why I put it in easy Wink
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rloginunix
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Re: Minimization problem  
« Reply #5 on: Jan 15th, 2015, 12:43pm »
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A substitution that gets rid of the radicals may be one approach.
 
I am not completely successful but bare with me and may be someone will find the one that works.
 
First term:
 
507 = 39*13
x2 - 39 x + 507 = x2 - 39(x - 13)
x - 13 = t
x = t + 13
x2 = t2 + 26t + 132
 
Put these back into the original expression for x:
 
t2 + 26t + 132 - 39t = t2 - 13t + 132
 
It is my wishful thinking of course but the above is 13t short of full square (t - 13)2.
 
Second term:
6348 = 138*46
y2 - 138y + 6348 = y2 - 138(y - 46)
y - 46 = v
y = v + 46
y2 = v2 + 92v + 462
 
which yields:
 
y2 - 138y + 6348 = v2 - 46v + 462
 
which is 46v short of full square (v - 46)2.
 
With a successful substitution I think the the third term will unfold nicely.
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Re: Minimization problem  
« Reply #6 on: Jan 16th, 2015, 1:34am »
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rloginunix: that's not what I did, but of course, it might work...
 
Smallish hint as to my approach: don't square roots of quadratic expressions remind you of something?
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Re: Minimization problem  
« Reply #7 on: Jan 16th, 2015, 2:19am »
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trigonometry?
« Last Edit: Jan 16th, 2015, 2:20am by towr » IP Logged

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Re: Minimization problem  
« Reply #8 on: Jan 16th, 2015, 2:29am »
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on Jan 16th, 2015, 2:19am, towr wrote:
trigonometry?

Yep.
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Re: Minimization problem  
« Reply #9 on: Jan 16th, 2015, 9:54am »
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This is interesting. Before my first post I've tried switching to polar coordinates as well as a number of trigonometric substitutions without success ...
 
x - 39/2 = (sqrt(507)/2)tan(t) and y - 69 = sqrt(1587)tan(v)?
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Re: Minimization problem  
« Reply #10 on: Jan 19th, 2015, 12:48pm »
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My answer does not involve any substitutions, neither polar nor trigonometric.
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rloginunix
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Re: Minimization problem  
« Reply #11 on: Jan 21st, 2015, 1:53pm »
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I think I got it - no need to take any derivatives whatsoever (nice puzzle, pex).
 
My previous number factorizations proved useful - I noticed that all the quadratics for each term can be rewritten in the law of cosines form:
 
x2 - 39x + 507 = x2 + (507)2 - 2x507 Cos(30)
 
y2 - 138x + 6348 = y2 + (6348)2 - 2y6348 Cos(30)
 
x2 - xy + y2 = x2 + y2 - 2xyCos(60)
 
Which means - we construct a triangle with two sides of 507 and 6348 and the angle of 30 + 30 + 60 = 120 degrees between them. The third side can be found easily. Next, x and y must be the lengths of the line segments on the "trisectors" of the 120-degree angle. Normally these line segments must either "stick out" or fall short of the third side of the triangle and the geometric interpretation of the given function is the sum of lengths of line segments from the vertexes of the base to the points X and Y each of which marks the length of the x and y line segment on the corresponding "trisector". But for minimization requirement they must fall dead on the base - smallest value = shortest distance in R3 = straight line. So the problem actually becomes of plane geometry.
 
I didn't run the numbers and didn't do the drawing yet, but you get the idea.
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Re: Minimization problem  
« Reply #12 on: Jan 21st, 2015, 5:08pm »
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Good job, rloginunix. Smiley It took me quite a bit of trial and error to get all nice and round numbers...
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Re: Minimization problem  
« Reply #13 on: Jan 21st, 2015, 7:20pm »
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You did a good job at that. I was drawing the triangles for three days, Smiley. The problem was - they were all different and I was hell bent on taking the derivatives. Only when I said "what will happen if I draw just one triangle?" it dawned on me ...
 
Anyway, there will be two 90-degree angles: between 507 and y and between 6348 and x. Which means that x and y are triangle's heights, we use square area and that's the second relationship between them. It's mostly arithmetic from here.
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