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.