Timeline for Is this min not less than a min
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2013 at 6:53 | comment | added | domotorp | I think all functions are differentiable in this problem. Or is your problem the max? It only takes the max of a few functions, so it should not cause a big problem. | |
| Mar 30, 2013 at 3:57 | comment | added | userior | @domotorp, can you show that the maximum in the second quantity at the case that $v_{0},v_{1},v_{2},v_{3}$and $v_{4}$ are all on the boundary of $\mbox{convexhull}\left(v_{0},v_{1},v_{2},v_{3},v_{4}\right)$ is greater than or equal to the maximum in the second quantity at the case of regular pentagon? Thanks a lot! | |
| Mar 30, 2013 at 3:18 | comment | added | userior | @domotorp, Lagrange multipliers normally requires the function to be minimized to be differentiable, but the function here to be minimized is not differentiabe at many places. | |
| Mar 30, 2013 at 2:53 | comment | added | userior | @Günter, I calculated both the value for the square plus its center and the square plus an optimal point inside the square, and the values are both greater than the extremal value in the case when all the five points are on the boundary. But do you know if the square plus an point inside realizes the the minimum over the case of one point inside a quadrilateral, or how to find the optimal quadrilateral? Thanks! | |
| Feb 27, 2013 at 16:32 | comment | added | Günter Rote | @userior, why don't you calculate the value for the square plus center and compare? Then at least we have a conjecture, and we only need to concentrate on one problem. | |
| Feb 26, 2013 at 17:03 | comment | added | domotorp | I think it would be the simplest to find the extreme configuration for that problem too but that seems harder, but probably not much. Maybe you can show by using Lagrange multipliers where the extreme value is. | |
| Feb 25, 2013 at 23:48 | comment | added | userior | domotorp , thanks for your answer, but the main difficulty in the question is how to show that the quantity for the case that one point inside a quadrilateral is greater than that for a regular pentagon. Do you know how to show that? Thanks! | |
| Feb 25, 2013 at 8:59 | history | answered | domotorp | CC BY-SA 3.0 |