.
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).
Solved.
To see more solved problems on the linear programming method, look into the lesson
- Solving minimax problems by the Linear Programming method
in this site.