SOLUTION: Maximize z = 2x + 4y subject to the constraints x &#8805; 0, y &#8805; 0, x + y &#8805; 1, 3x + 2y <= 6

 
When x is "greater than" it means "to the right of".
That means that the feasible region is on or to
the right of the line whose equation is  which
is the line which is the y-axis.

 
When y is "greater than" it means "above"
That means that the feasible region is on or 
above the line whose equation is  which
is the line which is the x-axis.

Those two mean the feasible region is in this upper right-
hand part of the xy-coordinate system:

 

 
If you solve that for x you get 
When x is "greater than" it means "to the right of"
If you solve that for y you get 
When y is "greater than" it means "above".
That means that the feasible region is on or 
above and to the right of the line whose equation is
.  The x-intercept of that line is (1,0),
and the y-intercept is (0,1). 

That's above and to the
right of the green line below, and of course right of
the y-axis and above the x-axis:

 

 
If you solve that for x you get 
When x is "less than" it means "to the left of"
If you solve that for y you get 
When y is "less than" it means "below".
That means that the feasible region is on or 
bolow and to the left of the line whose equation is
.  The x-intercept of that line is (2,0),
and the y-intercept is (0,3). That's below and to the 
left of the blue line below, and of course right of the 
y-axis and above the x-axis, and above and to the right 
of the green line:
 

The corner points of that region are (1,0), (2,0), (0,3), and (0,1)
So we list them in this table with the objective function evaluated
at each corner point:



Corner point    |  Value of z = 2x+4y
----------------|---------------------------                 
    (1,0)       |  z = 2(1)+4(0) = 2+0 = 2
                |
    (2,0)       |  z = 2(2)+4(0) = 4+0 = 4 
                |
    (0,3)       |  z = 2(0)+4(3) = 0+12 = 12
                |
    (0,1)       |  z = 2(0)+4(1) = 0+4 = 4

 
The instructions were to maximize the objective function.
So we see that the maximum value of the objective
function is 12 when x=0 and y=3.

[Note: If the problem had asked us to minimize
the objective function. the answer would have been that the minimum
value of the objective function would be 2 when x=1 and y=0.]


Edwin