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MathGroup Archive 2007

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Re: Is this a problem in mathematica? (2nd)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74621] Re: Is this a problem in mathematica? (2nd)
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Wed, 28 Mar 2007 04:35:22 -0500 (EST)
  • References: <eud3g1$l0t$1@smc.vnet.net>

Execute the following command also

FrontEndExecute[{HelpBrowserLookup["RegGuide", "Maximize"]}]

and see the "Furter Examples".

For your needs the following will give you waht you want.

(Minimize[(x + 0)^2 + (y - 2)^2, y == x^2 + 1 && #1[x, 0], {x, y}]
& ) /@ {Greater, Less}

{{3/4, {x -> 1/Sqrt[2], y -> 3/2}}, {3/4, {x -> -(1/Sqrt[2]), y ->
3/2}}}

Dimitris

=CF/=C7 traz =DD=E3=F1=E1=F8=E5:
> Let's say I wanna solve this problem:
>
> Determine point(s) on y = x^2 +1 that are
> closest to (0,2).
>
> So in mathematica:
>
> minDist = (x - 0)^2 + (y - 2)^2;
> Minimize[minDist, y == 1 + x^2, {x, y}]
>
> Output will give you: x -> -1/Sqrt[2], y -> 3/2
>
> but x also has another answer: +1/Sqrt[2]. Is this a problem in mathemati=
ca or can my code be changed to output the other value of x for the minimum=
 distance?



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