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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Mar 2008 20:33:46 IST
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x,y belongs R
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Umang |
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24 Mar 2008 20:38:17 IST
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is it -11??
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24 Mar 2008 20:40:42 IST
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yes it is
pls explain !
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Umang |
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24 Mar 2008 20:47:22 IST
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well c it lik this...wrt x y is const..
so lets call it f(x) where coeff include y...
now find f'(x)=0 u get rel x=3-y chk for f"(x) also
now v get tht @ x=3-y f(x) has min value= 2y^2+4y-9
(which u get by subst x in 1st eqn i.e f(x))
now v have to find min val of this...
so diff 2y^2 +4y-9=0 u get min value @ y=-1
so min val is -11
now if u r askin y not i initiall do as f(y) ok do the same way..
u get min value @ 3y=1-x in tht case
and if u follow same procedure u again get -11
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24 Mar 2008 20:51:30 IST
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its confusing !
pls explain in a simple manner !!
thank you !
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Umang |
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24 Mar 2008 20:52:44 IST
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welll...
u got 2 wait till some1 else comes along
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24 Mar 2008 21:14:29 IST
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hey !
i got it !
u hav done a lot of typing mistake !!!
thanks !!
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Umang |
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24 Mar 2008 21:15:26 IST
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now wait a minute...cud u point out a few???
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24 Mar 2008 21:19:08 IST
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@ x=3-y f(x)
u wrote @ instead of 2.
but anyways, thanks !!!
u appearing for JEE 2008 ?
got ur admit card ?
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Umang |
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24 Mar 2008 21:21:13 IST
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no mate its correct only...
this is wat i have said "now we get that at x=3-y , f(x) has min value"...
yeah i am...got my admit card a wk ago
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24 Mar 2008 21:29:17 IST
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oh sorry !
i understood !
now ur method is completely clear to me !!!
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Umang |
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25 Mar 2008 00:13:11 IST
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let us first fix the value of y and then vary x to find the minimum value of func. for that fixed value of y... we get x= 3-y.... so the min value for that fixed y is 2y^2 +4y -9..... we had kept y fixed......now we vary y too to get the least value among all values of y.... so we get at y= -1 , the expression has the min value for all x ,y........... thus min val= -11......
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