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MathGroup Archive 2007

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Re: Is this a problem in mathematica?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74668] Re: [mg74615] Is this a problem in mathematica?
  • From: Carl Woll <carlw at wolfram.com>
  • Date: Fri, 30 Mar 2007 03:01:32 -0500 (EST)
  • References: <200703280644.BAA20508@smc.vnet.net>

traz wrote:

>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 mathematica or can my code be changed to output the other value of x for the minimum distance?
>  
>
Another possibility besides those menioned by others is to use Solve 
once you know the minimum:

minDist=(x-0)^2+(y-2)^2;
Minimize[minDist,y==1+x^2,{x,y}]
{3/4,{x->-1/Sqrt[2],y->3/2}}

Solve[{minDist==3/4,y==1+x^2},{x,y}]
{{y->3/2,x->-1/Sqrt[2]},{y->3/2,x->-1/Sqrt[2]},{y->3/2,x->1/Sqrt[2]},{y->3/2,x->1/Sqrt[2]}}

Carl Woll
Wolfram Research


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