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 August 31st, 2013, 04:06 AM #1 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित minimum value of function minimum value of when subjected to and satisfies the condition. but I'm not able to show all three at a time I want someone to explain me the problem. August 31st, 2013, 05:46 AM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: minimum value of function Because of the cyclic symmetry of the variables, the minimum is found when: You will easily find this is true using Lagrange multipliers. August 31st, 2013, 08:11 AM #3 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Re: minimum value of function I can see , and has similar role, nothing happens if we interchange their position. could u please explain me in simple words. August 31st, 2013, 08:46 AM   #4
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Re: minimum value of function

Quote:
 Originally Posted by MATHEMATICIAN minimum value of when subjected to and satisfies the condition. but I'm not able to show all three at a time I want someone to explain me the problem.
I don't understand what you mean by "show all three at a time". I presume that "all three" refers to minimizing while satisfying x+ y+ z= 1 and xyz= -1. But what do you mean by "at a time"?

In any case, what have you tried? August 31st, 2013, 09:42 AM #5 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Re: minimum value of function what I did is, f = x + y + z . . . . (i) x + y + z - 1 = 0 . . . . (ii) xyz + 1 = 0 . . . . (iii) we know, x + y + z = ( x + y + z) - 2xy - 2yz - 2xz f = 1 - 2xy - 2yz - 2xz substituting value of "z" from equation (iii) f = 1 - 2xy + 2/x + 2/y . . . . (vi) partially differentiating equation (vi) w.r.t x and y and equation the results with zero, yx = - 1 . . . . (v) xy = - 1 . . . . (vi) combining (v) and (vi) yx - xy = 0 xy ( x - y ) = 0 thus, x = y why am i getting x = y if x = y, I get only (1, 1, -1) and don't get other point. is the process I followed incorrect ? August 31st, 2013, 09:47 AM #6 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Re: minimum value of function [color=#800040]Mark[/color] there are two conditions, x + y + z - 1 = 0 and xyz + 1 = 0 I don't know latex properly, so may b second condition is not displayed properly. August 31st, 2013, 10:08 AM   #7
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Re: minimum value of function

Quote:
 Originally Posted by MATHEMATICIAN [color=#800040]Mark[/color] there are two conditions, x + y + z - 1 = 0 and xyz + 1 = 0 I don't know latex properly, so may b second condition is not displayed properly.
Yes, I see that now after [color=#00BF00]HallsofIvy [/color]'s post. In your original post this was not clear. Lagrange multipliers here implies the system:

For which we find the real solutions:

and: August 31st, 2013, 10:21 AM #8 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Re: minimum value of function Could u please tell me someone about first three equations. also, mew and lamda. August 31st, 2013, 10:27 AM #9 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: minimum value of function I assumed that if you are studying optimization with constraints, then you are familiar with Lagrange multipliers. I simply do not have the patience with the quirky rendering of here to attempt a tutorial on the subject. August 31st, 2013, 10:33 AM #10 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Re: minimum value of function :'( Tags function, minimum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Dmath Calculus 4 March 9th, 2014 06:19 PM Fruit Bowl Applied Math 0 January 31st, 2013 10:25 AM david hilbert Complex Analysis 3 January 28th, 2013 02:33 AM bertusavius Calculus 1 October 18th, 2012 03:08 AM amateurmathlover Calculus 4 April 22nd, 2012 12:39 AM

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