# How to solve inequalities ?

Posted 9 years ago
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 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 & yBut,4x+2y<=440x+4y<=480x+y<=150x<=90 and of course x,y>=0How should I write it to solve my problem ?Thanks you.
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Posted 9 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 9 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