|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
(-∞,0)⋃(1,∞)
*******************************
|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 ≧ 1
y ≦ 2
y ≦ 3x+1
x ≦ 1
Draw the graph of y = 1 green. The feasible region is above the green line.
≦ ≧
Draw the line y = 2 in red. The feasible region is below the red line.
Draw the line y=3x+1 in black. The feasible reagion is below and to
the right of the black line:
Draw the line x = 1 in blue. The feasible reagion is to
the left of the blue line:
So, I'll erase everything except the feasible region,
which is this trapezoid, and I'll label 3 of the corner points already.
We'll have to calculate the upper left point by solving the system:
which gives the point (
,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
(,2) 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
