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Given $A$ being the intersection set between $x + y + z + u + v + w = 1$ and $x^2 + y^2 + z^2 + u^2 + v^2 = 1$, find the closest and farthest points to the origin that belong to $A$ in the space $\mathbb{R}^6$.

Is this even possible?

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  • $\begingroup$ Is this a question about the program Mathematica? If so, what have you tried thus far? $\endgroup$ Oct 13, 2019 at 14:37
  • $\begingroup$ Yes, Wolfram Mathematica 11.3 $\endgroup$
    – EichenH
    Oct 13, 2019 at 14:38
  • $\begingroup$ Besides what Neumann proposed, nothing too relevant. $\endgroup$
    – EichenH
    Oct 13, 2019 at 14:38
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    $\begingroup$ That being the case, this sort of question should first go to an instructor or TA for assistance in getting started. The scope of the forum starts, more or less, at the point where one presents actual code, along with specific questions about said code. $\endgroup$ Oct 13, 2019 at 14:53

1 Answer 1

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Try

NMinimize[{x^2 + y^2 + z^2 + u^2 + v^2 +w^2, {x + y + z + u + v + w == 1, x^2 + y^2 + z^2 + u^2 + v^2 == 1}}, {x, y, z, u, v, w}]
(*{1., {x -> -0.0316501, y -> -0.253188, z -> 0.544651, u -> 0.796892,v -> -0.0566812, w -> -0.0000230699}}*)

and

 NMaximize[{x^2 + y^2 + z^2 + u^2 + v^2 +w^2, {x + y + z + u + v + w == 1, x^2 + y^2 + z^2 + u^2 + v^2 == 1}}, {x, y, z, u, v, w}]
 (*{11.4721, {x -> -0.447214, y -> -0.447214, z -> -0.447214,u -> -0.447214, v -> -0.447214, w -> 3.23607}}*)
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  • $\begingroup$ Thanks for the idea! Shouldn't we try to define the intersection "A" first? And is the result plottable in any way? $\endgroup$
    – EichenH
    Oct 13, 2019 at 14:16
  • $\begingroup$ @DanielLichtblau - Thanks. I made a left turn in hyperspace and got lost. Deleted my comment. $\endgroup$
    – Bob Hanlon
    Oct 13, 2019 at 15:21
  • $\begingroup$ @BobHanlon Right. It ain't like dusting crops, boy $\endgroup$ Oct 13, 2019 at 15:25
  • $\begingroup$ @EichenH A is defined by the two constraints, I think. $\endgroup$ Oct 13, 2019 at 19:47

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