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The feasible region is presented in the plot below.
It is the region in the first quadrant above the given straight lines - the constraints.
The lines 2x+y = 30 (red), 2x+5y = 50 (green) and x+y = 20 (blue).
This region has vertices P1, P2, P3 and P4
P1 = (0,30), which is y-intercept of the red line 2x+y = 30;
P2 = (10,10), which is the intersection point of the red line 2x+y = 30 and blue line x+y = 20;
P3 = (, ), which is the intersection point of the blue line x+y = 20 and green line 2x+5y = 50;
P4 = (25,0), which is x-intercept of the green line 2x+5y = 50.
The graphic Linear Programming method includes finding the intersection points as the solutions of given constraint equations; these
procedures are considered as of atomic size components, and it is assumed that the student makes them automatically without special discussions.
Further, the Linear Programming method enact you to calculate the values of the objective function F(x,y) = 3x + 2y at the vertices
at P1: F( 0,30) = 3*0 + 2*30 = 60;
at P2: F(10,10) = 3*10 + 2*10 = 50;
at P3: F(,) = = ;
at P4: F(25,0) = 3*25 + 2*0 = 75.
Then the Linear Programming method STATES that the objective function in the feasible region achieves its minimum
at the vertex; namely at that vertex, where its value is minimal.
From the list above, the objective function has the minimum at P2 = (10,10), and this point is the minimum of the objective function
over the entire feasible region.
Thus the solution of the problem is the point P2 = (10,10).
It means that the objective function under the given constraints achieves its minimum at the point x = 10, y = 10.
The value of this minimum is = 50.
Answer. The objective function under the given constraints achieves its minimum at the point x = 10, y = 10.
The value of this minimum is = 50.
Solved.