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  Let x1,x2 & x3 be the quantity of rubber powder RA,RB & RC produced respectively.

  Maximise:Z=-2.05x1+100.6x2+40.85x3

  Subject to

  15x1+(25+2.2)x2+(12+5.1)x3<=60*6.5

  15x1+27.2x2+17.1x3<=390

  20x1+55x2<=5000

  45x1+100x2+72x3<=7500

  x1>=10,

  x2>=25,

  x3>=5

  Introducing slack,surplus & artificial variables

  Maximise:Z=-2.05x1+100.6x2+40.85x3+0S1+0S2+0S3+0S4+0S5+0S6-MA1-MA2-MA3

  Subject to

  15x1+27.2x2+17.1x3+S1<=390

  20x1+55x2+S2<=5000

  45x1+100x2+72x3+S3<=7500

  x1-S4+A1>=10,

  x2-S5+A2>=25,

  x3-S6+A3>=5

  x1,x2,x3,S1,S2,S3,S4,S5,S6,A1,A2,A3>=0

  Negative minimum Zj-Cj is -M-100.6 and its column index is 2. So, the entering variable is x2.
  
  Minimum ratio is 14.3382 and its row index is 1. So, the leaving basis variable is S1.
  
  ∴ The pivot element is 27.2.
  
  Entering =x2, Departing =S1, Key Element =27.2

  

  Negative minimum Zj-Cj is -0.3713M+22.3949 and its column index is 3. So, the entering variable is x3.
  
  Minimum ratio is 5 and its row index is 6. So, the leaving basis variable is A3.Negative minimum Zj-Cj is -0.3713M+22.3949 and its column index is 3. So, the entering variable is x3.
  
  Minimum ratio is 5 and its row index is 6. So, the leaving basis variable is A3.
  
  ∴ The pivot element is 1.

  Since all Zj-Cj≥0
  
  Hence, optimal solution is arrived with value of variables as :
  x1=10,x2=5.6801,x3=5
  
  Max Z=755.1728

  So to obtain maximum profit value of constraint x2>=25 should be changed to x2>=5.

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