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I want to find the limit value (range) of "x", "y" and "z" on which above above function is positive for given range. for this i use

RegionPlot3D[x^2 - y + z - x y z < 0, {x, 1, 10}, {y, -10, 10}, {z, 1, 10}]

where f[x_ ,y_ ,z_]:= x^2 - y + z - x * y * z

now how to find the limit value of x,y,z from this plot??? any other method???

See https://i.sstatic.net/MJ23M.png

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3 Answers 3

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This answers the question of how to get "limit values from RegionPlot3D?"

RegionBounds@ DiscretizeGraphics@ RegionPlot3D[x^2 - y + z - x y z < 0,
    {x, 1, 10}, {y, -10, 10}, {z, 1, 10}]
(*  {{1., 10.}, {0.613142, 10.}, {1., 10.}}  *)

Of course, RegionPlot3D makes a rough, discrete approximation to the actual region, so some numerical error is present.

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This can be done as follows.

Maximize[y, x^2 - y + z - x y z < 0 && x >= 1 && x <= 10 && y >= -10 && y <= 10 &&z >= 1 && z <= 10, {x, y, z}]

{10, {x -> 6, y -> 10, z -> 6}}

Minimize[y,x^2 - y + z - x y z < 0 && x >= 1 && x <= 10 && y >= -10 && y <= 10 &&   z >= 1 && z <= 10, {x, y, z}]

{1/50 (-1 + Sqrt[1001]), {x -> 1/10 (-1 + Sqrt[1001]) - Sqrt[-10 + 1/50 (-1 + Sqrt[1001]) + 1/100 (-1 + Sqrt[1001])^2], y -> 1/50 (-1 + Sqrt[1001]), z -> (1/50 (1 - Sqrt[1001]) + (1/10 (-1 + Sqrt[1001]) - Sqrt[-10 + 1/50 (-1 + Sqrt[1001]) + 1/100 (-1 + Sqrt[1001])^2])^2)/(-1 + 1/50 (-1 + Sqrt[1001]) (1/10 (-1 + Sqrt[1001]) - Sqrt[-10 + 1/50 (-1 + Sqrt[1001]) + 1/100 (-1 + Sqrt[1001])^2]))}}

The bounds of $x$ and $z$ can be found similarly.

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Maximize[y, x^2 - y + z - x y z < 0 && x >= 1 && x <= 10 && y >= -10 && y <= 10 &&z >= 1 && z <= 10, {x, y, z}]

and

Minimize[y, x^2 - y + z - x y z < 0 && x >= 1 && x <= 10 && y >= -10 && y <= 10 &&z >= 1 && z <= 10, {x, y, z}]

is not the range but the boundary value (upper/lower) and is not the complete limit range.

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  • $\begingroup$ @ Steve: Can you base your statement? Of course, by optimization the end points of the ranges under consideration are obtained. $\endgroup$
    – user64494
    Oct 25, 2016 at 18:22
  • $\begingroup$ any other method?? $\endgroup$
    – Steve
    Oct 25, 2016 at 18:32
  • $\begingroup$ @ Steve. Sorry, don't understand you. I will be waiting for a reply. $\endgroup$
    – user64494
    Oct 25, 2016 at 18:37

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