max = x 1+x2；

2 * x 1+5 * x2 & lt； = 16 ;

6 * x 1+5 * x2 & lt； =30 ;

x 1 & gt； =0 ;

x2 & gt=0;

@ free(x2)；

According to the operation, the optimal solution of the model is obtained as follows:

Target value: 5.300000

Variable valve reduces cost

x 1 3.500000 0.000000

x2 1.8000000 1.000000