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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
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