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 MATHEMATICIAN August 31st, 2013 04:06 AM

minimum value of function

minimum value of $x^2 + y^2 + z^2$
when subjected to $x + y + z -1= 0$ and $xyz + 1= 0$

$(1, 1, -1), (1, -1,1) & (-1, 1, 1)$ satisfies the condition.
but I'm not able to show all three at a time
I want someone to explain me the problem.

 MarkFL August 31st, 2013 05:46 AM

Re: minimum value of function

Because of the cyclic symmetry of the variables, the minimum is found when:

$x=y=z=\frac{1}{3}$

You will easily find this is true using Lagrange multipliers.

 MATHEMATICIAN August 31st, 2013 08:11 AM

Re: minimum value of function

I can see $x$, $y$ and $z$ has similar role, nothing happens if we interchange their position.

could u please explain me in simple words.

 HallsofIvy August 31st, 2013 08:46 AM

Re: minimum value of function

Quote:
 Originally Posted by MATHEMATICIAN minimum value of $x^2 + y^2 + z^2$ when subjected to $x + y + z -1= 0$ and $xyz + 1= 0$ $(1, 1, -1), (1, -1,1) & (-1, 1, 1)$ 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 $x^2+ y^2= z^2$ 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?

 MATHEMATICIAN August 31st, 2013 09:42 AM

Re: minimum value of function

what I did is,

f = x$^2$ + y$^2$ + z$^2$ . . . . (i)

x + y + z - 1 = 0 . . . . (ii)

xyz + 1 = 0 . . . . (iii)

we know,
x$^2$ + y$^2$ + z$^2$ = ( x + y + z)$^2$ - 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$^2$ = - 1 . . . . (v)

xy$^2$ = - 1 . . . . (vi)

combining (v) and (vi)
yx$^2$ - xy$^2$ = 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 ?

 MATHEMATICIAN August 31st, 2013 09:47 AM

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.

 MarkFL August 31st, 2013 10:08 AM

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:

$2x=\lambda+\mu yz$

$2y=\lambda+\mu xz$

$2z=\lambda+\mu xy$

$x+y+z-1=0$

$xyz\,+\,1\,=\,0$

For which we find the real solutions:

$(\lambda,\mu)=(0,-2)$

and:

$(x,y,z)=(-1,1,1),\,(1,-1,1),\,(1,1,-1)$

 MATHEMATICIAN August 31st, 2013 10:21 AM

Re: minimum value of function

Could u please tell me someone about first three equations.

also, mew and lamda.

 MarkFL August 31st, 2013 10:27 AM

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 $\LaTeX$ here to attempt a tutorial on the subject.

 MATHEMATICIAN August 31st, 2013 10:33 AM

Re: minimum value of function

:'(

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