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?