Timeline for How to plot all the points which satisfy equation/inequality in two variables?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2019 at 3:36 | vote | accept | Shirin elahi | ||
| Jul 29, 2019 at 21:16 | comment | added | Shirin elahi | Thanks Bill and "user64494". RegionPlot worked perfectly | |
| Jul 29, 2019 at 17:40 | comment | added | user64494 | Have you tried RegionPlot[ y > 11 + (1 + 2 x) ((11 x)/(5 + 10 x))^((3 y)/(3 y - (5 + x))), {x, 0, 1000}, {y, 0, 1000}] which works well in version 12.0? | |
| Jul 29, 2019 at 16:45 | comment | added | Bill |
Your second question appears to puzzle Mathematica, but it appears that if you choose any of those {x,y} values returned from FindInstance then every y value greater than that will satisfy your second question. Thus N[FindInstance[y==11+(1+2x)((11x)/(5+10x))^((3y)/(3y-(5+x)))&&x>0&&y>0,{x,y}]] showed me {{x->4.10529,y->20.}} and x==4.10529,y==20.01 or y==20.02 or y=30 or y=1000000 all seem to satisfy your y>11+(1+2x)((11x)/(5+10x))^((3y)/(3y-(5+x))) condition. Does this work for you?
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| Jul 29, 2019 at 16:25 | vote | accept | Shirin elahi | ||
| Jul 29, 2019 at 21:12 | |||||
| Jul 29, 2019 at 16:19 | comment | added | Shirin elahi | Thanks Bill. It works perfectly. I think I am gonna need to solve the inequality to find a range for y. May I ask if you have any idea for the second question? | |
| Jul 29, 2019 at 16:13 | history | answered | Bill | CC BY-SA 4.0 |