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| Question 1142000:  Solve the linear programming problem by the method of corners.
 Maximize	P = x + 6y
 subject to x	+ y	≤	4
 2x	+ y	≤	7
 x ≥ 0, y ≥ 0
 The maximum is P =
 at
 (x, y) =
 
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . 
 
You need to maximize the objective function P(x,y) = x + 6y under given restrictions.
The feasible domain is shown below.
It is  a quadrilateral in the first quadrant  (x >= 0,  y >= 0)  restricted 
by the red line  x + y = 4  and the green line  2x + y = 7.
 Plots  x + y = 4  (red) and  2x + y = 7 (green)
The vertices of this quadrilateral are
        (X0,Y0) = (0,0)     (the origin of the coordinate system);
        (X1,Y1) = (0,4)     (red line Y-intercept);
        (X2,Y2) = (3,1)     (intersection point of the straight lines y = 4-x and  y = 7-2x );
        (X3,Y3) = (3.5,0)   (green line X-intercept)
Calculate the objective function at these points
        P(X0,Y0) = 0 + 6*0   = 0;
        P(X1,Y1) = 0 + 6*4   = 24;
        P(X2,Y2) = 3 + 6*1   =  9;
        P(X3,Y3) = 3.5 + 6*0 =  3.5.
Select one of these points where the objective function is maximal. In our case this point is (X1,Y1) = (0,4).
This point gives your optimal solution x = 0,  y = 4.
The maximum objective function value is 24. Solved.
 
 ------------------
 
 To see many other similar solved problems, look into the lesson
 - Solving minimax problems by the Linear Programming method
 in this site.
 
 See also many other similar solved problems in the archive to this forum
 
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 From these sources, learn  the technique and the methodology on how to solve minimax problems using the Linear Programming method - once and for all.
 
 
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