SOLUTION: Solve the linear programming problem. Minimize z = 11x + 6y + 7 subject to: x ≥ 0, y ≥ 0, x + y ≥ 1.

For this problem, the area of feasibility is the triangle in a coordinate plane in the first quadrant, restricted by 
the x-axis, y-axis and the straight line x + y = 1. 


The vertices of this triangle are the points (0,0), (0,1)  and  (1,0).


According to the linear programming method, in order to minimize z = 11x + 6y + 7,

    you need to calculate the function z(x,y) in the vertices of this triangle, to compare the results and to choose 
    that of the three points where the z(x,y) is minimal.


1)  Point (0,0):  z(0,0) = 11*0 + 6*0 + 7 = 7,


2)  Point (0,1):  z(0,1) = 11*0 + 6*1 + 7 = 6 + 7 = 13,


2)  Point (1,0):  z(1,0) = 11*1 + 6*0 + 7 = 11 + 7 = 18.



So, the minimum value of z(x,y) is achieved at the point (0,0).


Then the linear programming method states that the point (0,0) gives the minimum to the given linear function z(x,y) = 11x + 6y + 7 over the triangle.


Answer.  Under the given restrictions, the linear function z(x,y) = 11x + 6y + z is minimal at the point (0,0).