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# How to solve inequalities ?

Posted 10 years ago
 Hi everyone, I'm a new user of Wolfram and I want to know how to solve some inequalities. I spend a lot of time for an easy problem but I don't know how to write it in terms of wolfram language. So, I need your help. Here's my problem : z=100x+200y to find z i have to find x & y But, 4x+2y<=440 x+4y<=480 x+y<=150 x<=90 and of course x,y>=0 How should I write it to solve my problem ? Thanks you.
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Posted 10 years ago
 Oh great, thanks you. Wolfram advised me to reduce it but i didn't know how to do it. Need to spend some time on this new program.
Posted 10 years ago
 Well, you could look for the interval for which all the conditions on x and y are fulfilled. This can be done like so: Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 &&x >= 0 && y >= 0, {x, y}] You get (0 <= x <= 40 && 0 <= y <= (480 - x)/4) || (40 < x <= 70 && 0 <= y <= 150 - x) || (70 < x <= 90 && 0 <= y <= 220 - 2 x) You can get z Solve[z == 100 x + 200 y && (Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}]), z, Reals] which gives: {{z -> ConditionalExpression[100 x + 200 y, (0 < x < 90 && 0 < y < 40) || (x > 0 && 40 < y < 80 && -220 + 2 x + y < 0) || (x > 0 && 80 < y < 110 &&x + y < 150) || (x > 0 && 110 < y < 120 && x + 4 y < 480)]}} If you want to maximize z you might want to look into expressions like: Maximize[100 x + 200 y , (Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}]), {x, y}] which finds: {26000, {x -> 40, y -> 110}} If you feel like it, you can also plot the whole thing: Show[Plot3D[100 x + 200 y, {x, 0, 200}, {y, 0, 200}, RegionFunction -> Function[{x, y, z}, Evaluate[ Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}][[1]]]]], Plot3D[100 x + 200 y, {x, 0, 200}, {y, 0, 200}, RegionFunction -> Function[{x, y, z}, Evaluate[ Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}][[2]]]]], Plot3D[100 x + 200 y, {x, 0, 200}, {y, 0, 200}, RegionFunction -> Function[{x, y, z}, Evaluate[ Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}][[3]]]]]] If you minimize the expression: Minimize[100 x + 200 y , (Reduce[4 x + 2 y <= 440 && x + 4 y <= 480 && x + y <= 150 && x <= 90 && x >= 0 && y >= 0, {x, y}]), {x, y}] you find {0, {x -> 0, y -> 0} so it appears that z can take values between 0 and 26000.Cheers,Marco