Question 473756
<pre>
|4x-2| > 2  becomes

4x-2<-2 OR 4x-2>2
  4x<0  OR   4x>4
   x<0  OR    x>1

<==========o----o==========>
-2   -1    0    1    2    3 

(-&#8734;,0)&#8899;(1,&#8734;)

*******************************

|2x-4| < 2 becomes:

    -2 < 2x-4 < 2
    +4     +4  +4
   ---------------
     2 < 2x   < 6
     1 <  x   < 3

-----------o=========o----------
-1    0    1    2    3    4    5 

             (1,3)



-----------------------------------------------
Maximize P=3x-y subject to:
y &#8807; 1
y &#8806; 2
y &#8806; 3x+1
x &#8806; 1

Draw the graph of y = 1 green. The feasible region is above the green line.

{{{drawing(2000/7,400,-.5,2,-.5,3,
graph(2000/7,400,-.5,3,-.5,3), green(line(-5,1,5,1)) )}}}

&#8806; &#8807;

Draw the line y = 2 in red. The feasible region is below the red line.


{{{drawing(2000/7,400,-.5,2,-.5,3, red(line(5,2,-5,2)),
graph(2000/7,400,-.5,2,-.5,3), green(line(-5,1,5,1)) )}}}

Draw the line y=3x+1 in black. The feasible reagion is below and to
the right of the black line:

{{{drawing(2000/7,400,-.5,2,-.5,3, red(line(5,2,-5,2)), line(6,19,-6,-17),
graph(2000/7,400,-.5,2,-.5,3), green(line(-5,1,5,1)) )}}}


Draw the line x = 1 in blue. The feasible reagion is to
the left of the blue line:

{{{drawing(2000/7,400,-.5,2,-.5,3, red(line(5,2,-5,2)), line(6,19,-6,-17),
graph(2000/7,400,-.5,2,-.5,3), green(line(-5,1,5,1)), blue(line(1,-6,1,6)) )}}}

So, I'll erase everything except the feasible region,
which is this trapezoid, and I'll label 3 of the corner points already.

{{{drawing(2000/7,400,-.5,2,-.5,3, red(line(1,2,1/3,2)), line(1/3,2,0,1),
graph(2000/7,400,-.5,2,-.5,3), green(line(0,1,1,1)), blue(line(1,1,1,2)), 
locate(0,1,"(0,1)"), locate(1,1,"(1,1)"), locate(1,2,"(1,2)"))}}}

We'll have to calculate the upper left point by solving the system:

{{{system(y=3x+1,y=2)}}}

which gives the point ({{{1/3}}},2)

{{{drawing(2000/7,400,-.5,2,-.5,3, red(line(1,2,1/3,2)), line(1/3,2,0,1),
graph(2000/7,400,-.5,2,-.5,3), green(line(0,1,1,1)), blue(line(1,1,1,2)), 
locate(0,1,"(0,1)"), locate(1,1,"(1,1)"), locate(1,2,"(1,2)"),
locate(-.05,2.3,nu(1/3,2))
)}}}

Now to maximize P = 3x-y.  We know the maximum (and minimum)
value must be at one of the 4 corner points, so we make this chart:

CORNER
POINT    3x  - y =          P
(0,1)   3(0) - 1 = 0 - 1 = -1
(1,1)   3(1) - 1 = 3 - 1 =  2   
(1,2)   3(1) - 2 = 3 - 2 =  1
({{{1/3}}},2)  3({{{1/3}}}) - 2 = 1 - 2 = -1

So the maximum value of P is 2 which occurs at 
the point (1,1) where x = 1 and y = 1.

Edwin</pre>