SOLUTION: 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) =

Algebra ->  Inequalities -> SOLUTION: 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) =       Log On


   



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(52786) About Me  (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.



graph%28+330%2C+330%2C+-2%2C+10%2C+-2%2C+10%2C%0D%0A++++++++++4-x%2C++7-2x%0D%0A%29


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.

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