Sistema inecuaciones con dos incógnitas, pares (x,y) zona factible

I'm sure that there will be a more compact way to do it.

Anyway, just to avoid bothering with nested commands:

- create the first line and find its intersections A and B with axes, then do the same with the other line (C, D)

- create a list containing integer coordinates 0<=x<=floor(x(A)) and 0<=y<=floor(y(B)) and do the same with the other line

- keep only the points in each list that are ok with the given inequality (see my "reds" and "blues" lists)

- count how many elements they contain ("redcount" and "bluecount")

- intersect the "reds" and "blues" lists

- count how many points are in the intersection (feasible area)

Many points! So the file is slow. This can be done also with more compact commands and a less step-by-step approach.

Anyway... check the attached file.

Thank you Simona. Very good hard work.

Hi Simona!, good evening. I have read the inside part of your work [I mean the instructions] and find them difficult to understand due to my insufficient GG knowledge. Moreover, it is also a problem to find good videos in the Internet about GG (at least in Spanish). Maybe might you help me on that?

Thank you once more for your help.

Jose Antonio.

si tus cuatro inecuaciones se llaman a,b,c,d (por ejemplo d es d:y>=0) define la inecuacion global a&&b&&c&&d (el doble simbolo && equivale al y logico)

supongamos que esta nueva inecuacion global se llama m (lo más probable es que se llame e) entonces usa la instruccion en ingles keepif(m(P),P,{vertex(m)}) copiala porque es facil equivocar las llaves y los parentesis

luego con las instrucciones de lista para maximo minimo e indexof puedes buscar el punto concreto

Muchas gracias mathmagic. Acabo de ver tu post. Intentaré implementar lo que me dices. Te mantendré informado.