Using results of Reduce

Question

Suppose I have function f:

f[x_, y_] := 50000 + x 30000 + y 35000;

Now I want to find the x and y when f[]<=200000. I use Reduce:

Reduce[f[x, y] <= 200000 && 1 <= x <= 8 && 1 <= y <= 8 , {x,y}, Integers]

The result is:

(x == 1 && y == 1) || (x == 1 && y == 2) || (x == 1 && y == 3) || 
(x == 2 && y == 1) || (x == 2 && y == 2) || (x == 3 && y == 1)

Now I would like to use these results back into f[] but it requires formatting. I use:

results = Reduce[f[x, y] <= 200000 && 1 <= x <= 8 && 1 <= y <= 8, 
           {x, y},Integers] /. 
             Or[a__, b__] -> {a, b} /. 
              And[a_ == c_, b_ == d_] -> {c, d}

And I use:

f[#[[1]], #[[2]]] & /@ results

to get:

{115000, 150000, 185000, 145000, 180000, 175000}

I wonder if this is the right way to do this or if it can be done easier. It seems reduce can generate different (depending on the equations of course) formats of the results and this would mean that for every type of result new rules will be needed. Is this a fact of life?

Answer 1

You may want to look at the documentation for ToRules. When applied to the output of Reduce you provided, the result is

Sequence[{x -> 1, y -> 1}, {x -> 1, y -> 2}, {x -> 1, y -> 3}, 
         {x -> 2, y -> 1}, {x -> 2, y -> 2}, {x -> 3, y -> 1}]

If you want to substitute this back to f[x,y], put curly braces to turn the sequence into a list and use ReplaceAll:

f[x,y] /. {%}

This will give the desired output {115000, 150000, 185000, 145000, 180000, 175000}.

Answer 2

I'd do the following:

res = Reduce[f[x, y] <= 200000 && 1 <= x <= 8 && 1 <= y <= 8, {x, y}, Integers];
vals = Apply[List, res[[All, All, 2]], {0, 1}]
(*
==> {{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {3, 1}}
*)

which uses Part in the form res[[All,All,2]] to get only the numerical values, and then Apply to replace the Or and And in the logical construct with List. For the second part, I'd again use Apply in its second short form:

f @@@ vals

Answer 3

50000 + x*30000 + y*35000 /. FindInstance[50000 + x*30000 + y*35000 < 200000 && 
    ((x == 1 && y == 1) || (x == 1 && y == 2) || (x == 1 && y == 3) || 
     (x == 2 && y == 1) || (x == 2 && y == 2) || (x == 3 && y == 1)), {x, y}, 
   Integers, 6]

(* ==>

{115000, 150000, 185000, 145000, 180000, 175000}

*)