Question 251750
Maximize z = 2x + 4y subject to the constraints

x &#8805; 0, y &#8805; 0, x + y &#8805; 1, 3x + 2y <= 6

<pre><font size = 4 color = "indigo"><b>
{{{x>=0}}} 
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 {{{x=0}}} which
is the line which is the y-axis.

{{{y>=0}}} 
When y is "greater than" it means "above"
That means that the feasible region is on or 
above the line whose equation is {{{y=0}}} 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:

{{{graph(400,400,-.2,3.5,-.2,3.5)}}} 

{{{x+y>=1}}} 
If you solve that for x you get {{{x>=1-y}}}
When x is "greater than" it means "to the right of"
If you solve that for y you get {{{y>=1-x}}}
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
{{{x+y=1}}}.  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:

{{{drawing(400,400,-.2,3.5,-.2,3.5,-2,

graph(400,400,-.2,3.5,-.2,3.5,-2), green(line(0,1,1,0)) )}}} 

{{{3x + 2y <= 6}}} 
If you solve that for x you get {{{x<=(6-2y)/3}}}
When x is "less than" it means "to the left of"
If you solve that for y you get {{{y<=(6-3x)/2}}}
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
{{{3x+2y=6}}}.  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:
{{{drawing(400,400,-.2,3.5,-.2,3.5,-2,
graph(400,400,-.2,3.5,-.2,3.5,-2), green(line(0,1,1,0)),
blue(line(2,0,0,3)), locate(.1,1.5,"feasible"), locate(.3,1.35,"region")
 )}}} 

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:

{{{z = 2x + 4y}}}

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.

<i><font size = 1>[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.]
</i></font>

Edwin</pre></font></b>