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Home > Operation Research calculators > Graphical method calculator
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Method
Solve the Linear programming problem using
Graphical method calculator
Type your linear programming problem
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Total Variables : Total Constraints :
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Steps using `Z_j-C_j` or `C_j-Z_j` :
  1. max z = -2x1 - x2
    subject to
    -3x1 - x2 <= -3
    -4x1 - 3x2 <= -6
    -x1 - 2x2 <= -3
    and x1,x2 >= 0
  2. max z = 15x1 + 10x2
    subject to
    4x1 + 6x2 <= 360
    3x1 <= 180
    5x2 <= 200
    and x1,x2 >= 0
  3. max z = 2x1 + x2
    subject to
    x1 + 2x2 <= 10
    x1 + x2 <= 6
    x1 - x2 <= 2
    x1 - 2x2 <= 1
    and x1,x2 >= 0
  4. max z = -x1 + 2x2
    subject to
    x1 - x2 <= -1
    -0.5x1 + x2 <= 2
    and x1,x2 >= 0
  5. max z = 40x1 + 80x2
    subject to
    2x1 + 3x2 <= 48
    x1 <= 15
    x2 <= 10
    and x1,x2 >= 0
  6. max z = 60x1 + 40x2
    subject to
    x1 <= 25
    x2 <= 35
    2x1 + x2 <= 60
    and x1,x2 >= 0
  7. min z = 3x1 + 2x2
    subject to
    5x1 + x2 >= 10
    x1 + x2 >= 6
    x1 + 4x2 >= 12
    and x1,x2 >= 0
  8. min z = 600x1 + 400x2
    subject to
    3x1 + 3x2 >= 40
    3x1 + x2 >= 40
    2x1 + 5x2 >= 44
    and x1,x2 >= 0
  9. min z = 4x1 + 3x2
    subject to
    200x1 + 100x2 >= 4000
    x1 + 2x2 >= 50
    40x1 + 40x2 >= 1400
    and x1,x2 >= 0
  10. min z = -x1 + 2x2
    subject to
    -x1 + 3x2 <= 10
    x1 + x2 <= 6
    x1 - x2 <= 2
    and x1,x2 >= 0
  11. max z = -15x1 - 10x2
    subject to
    -3x1 - 5x2 <= -5
    -5x1 - 2x2 <= -3
    and x1,x2 >= 0
  12. max z = 600x1 + 500x2
    subject to
    2x1 + x2 >= 80
    x1 + 2x2 >= 60
    and x1,x2 >= 0
  13. max z = 3x1 + 2x2
    subject to
    x1 - x2 >= 1
    x1 + x2 >= 3
    and x1,x2 >= 0
  14. max z = 5x1 + 4x2
    subject to
    x1 - 2x2 <= 1
    x1 + 2x2 >= 3
    and x1,x2 >= 0
  15. max z = -4x1 + 3x2
    subject to
    x1 - x2 <= 0
    x1 <= 4
    and x1,x2 >= 0
  16. max z = 3x1 + 4x2
    subject to
    x1 - x2 = -1
    -x1 + x2 <= 0
    and x1,x2 >= 0
  17. max z = 6x1 - 4x2
    subject to
    2x1 + 4x2 <= 4
    4x1 + 8x2 >= 16
    and x1,x2 >= 0
  18. max z = x1 + 1/2x2
    subject to
    3x1 + 2x2 <= 12
    5x1 = 10
    x1 + x2 >= 8
    -x1 + x2 >= 4
    and x1,x2 >= 0
  19. max z = 3x1 + 2x2
    subject to
    -2x1 + 3x2 <= 9
    3x1 - 2x2 <= -20
    and x1,x2 >= 0
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