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

*To*: mathgroup at smc.vnet.net*Subject*: [mg74631] Re: Is this a problem in mathematica? (2nd)*From*: Peter Pein <petsie at dordos.net>*Date*: Thu, 29 Mar 2007 02:29:01 -0500 (EST)*References*: <eud3g1$l0t$1@smc.vnet.net> <eudcub$s6o$1@smc.vnet.net>

dimitris schrieb: > 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? > > Or remember that the derivative of the distance has to be zero and the second (or any even) derivative has to be >0 at the minima: In[1]:= distsquare[x_] := (#1 . #1 & )[{x, 1 + x^2} - {0, 2}]; Reduce[distsquare'[x] == 0 && distsquare''[x] > 0, x] Out[2]= x == -(1/Sqrt[2]) || x == 1/Sqrt[2] P²