Solving linear programming problems with MATLAB and lingo

Production plan for processed dairy products
Processing 1 barrels of milk has two processing methods, (1) Production 12 hours to get 3 kg A1, profit 24 Yuan/kg (2) Production 8 hours to get 4 kg A2,

Profit 16 Yuan/kg
Every day:
50 barrels of milk time 480 hours processing up to 100 kg A1
Develop production plans to maximize daily profit

Analysis:
Set x1 barrels milk production a1x1 barrels milk production A2
The production of A1 profit 24X3X1 production A2 profit 16x4 x2
Daily profit
Max z=72x1+64x2
Constraint conditions
Raw Material x1+x2≤50
Time 12xx1+8xx2≤480
Processing capacity 3xx1≤100
Non-negative constrained x1,x2≥0
Lingo Code

Model:
max=72*x1+64*x2;
[Milk] x1+x2<50;
[Time]
12*x1+8*x2<480;
[cpct] 3*x1<100;
End

Results

  Global optimal solution found.
  Objective value:                              3360.000
  infeasibilities:                              0.000000 Total
  solver iterations:                             2


                       Variable           Value        Reduced cost
                             X1        20.00000            0.000000
                             X2        30.00000            0.000000

                            Row    Slack or Surplus      Dual Price
                              1        3360.000            1.000000
                           MILK        0.000000            48.00000
                           time        0.000000            2.000000
                           cpct        40.00000            0.000000

MATLAB solution

f= (-1) *[72 64]; % because the maximum value is required here to take the opposite number, Linprog is to find the minimum value of
a=[1 1;12 8;3 0;-1 0;0-1];
B=[50 480 0 0];
[Xopt Fxopt]=linprog (f,a,b)
-fxopt   % and then take negative to get maximum value

Code 2

f= (-1) *[72 64]; % because the maximum value is required here to take the opposite number, Linprog is to find the minimum value of
a=[1 1;12 8;3 0];
B=[50 480];
LB=[0;0];
[Xopt Fxopt]=linprog (f,a,b,[],[],lb)
-fxopt   % and then take negative to get maximum value

Results

Optimization terminated.

xopt =

   20.0000
   30.0000


fxopt =

 -3.3600e+003


ans =

  3.3600e+003

General form of linear programming
Minf=c1x1+c2x2+...+cnxn \min F=c_1x_1+c_2x_2+...+c_nx_n
S.t.⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪a11x1+a12x2+...+a1nxn≤b1a21x1+a22x2+...+a2nxn≤b2..............................am1x1+a