# Linear Equation by nkc8ft

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```									Maximum and Minimum
Chapter 6

1
Find inequalities from region
In the graph, find inequalities for regions A, B, C, D, E

2
4 x  3 y  24

Region A has 3 boundaries  x  2 y  10
     x0

So it is
4 x  3 y 24

 x  2 y 10

     x 0

3
 x  2 y  10
4 x  3 y  24


Region B has 5 boundaries  x  y  4
     x0


     y0
So it is
 x  2 y 10
4 x  3 y 24


 x y 4
     x 0


     y 0

4
 x  2 y  10
4 x  3 y  24

Region C has 4 boundaries 
 x y 4

     x0
So it is
 x  2 y 10
4 x  3 y 24


 x y 4
 x 0


5
 x  2 y  10

Region D has 3 boundaries 4 x  3 y  24
 x y 4

So it is
 x  2 y 10

4 x  3 y 24
 x y 4


6
 x y 4

Region E has 3 boundaries 4 x  3 y  24
 y0

So it is
 x y 4

4 x  3 y 24
 y 0


7
Exercises
1. In the graph, find inequalities for regions F and G

8
Positive linear functions
If a  0, b  0 then z  ax  by is called a positive linear function.
If z  c then the line ax  by  c is called a z-value line.
1) All z-value lines are parallel.
2) Value of z  ax  by is increasing from southwest to northeast
For example z  2 x  3 y, draw 0-value, 6-value and  6  -value lines.

9
z-value lins of z  2 x  3 y

10
Maximum and Minimum
In a finite polygon region, any positive linear function z  ax  by will
reach its maximum and minimum values at the corner points.

Maximum value point

Minimum value point

11
No maximum in unbounded region
In a unbounded region (of Quadrant I), a positive linear function z  ax  by
only has the minimum value, but NO MAXIMUM

No Maximum value

Minimum value point

12
Example
Find maximum value of the subject function z  3x  4 y in the region
 2x  y  4
 x  2 y  4


 x0
 y0

First we draw 4 lines
 2x  y  4
 x  2 y  4


 x0
 y0

Then we shadow the region

13
Next we have to decide the corner points. Among 4 corner points,
3 of them are already known. They are (0,0), (0,2) and (2,0).
We need to solve the last one, which is the common point of lines
2 x  y  4 and  x  2 y  4
Therefore , we setup
 2 x  y  4    (1)

 x  2 y  4    (2)
(1)  2(2) to get 5 y  12
12
y            (3)
5
Plug into (2)
12               4
x  2       4 x 
5               5

Corner points are (0,0), (0,2), (2, 0) and    
4 12
,
5 5
14
Now we calculate value of z  3x  4 y at those 4 corner points and
draw the following table

Corner
Value z = 3x + 4y
points
(0, 0)                      0
(0, 2)              30+42 = 8
(2, 0)              32+40 = 6

 
4 12
,
5 5
3
4
5
 4
12
5
 12 Maximum

The maximum value of z is 12 at point              4 12
,
5 5    
15
Example
Find maximum and minimum values of z  x  10 y in the region
 x  4 y  12
 x  2y  0


 2y  x  6
 x6

First we draw 4 lines
 x  4 y  12
 x  2y  0


 2y  x  6
 x6

Then we shadow the region

16
Next we have to decide the corner points. Among 4 corner points,
3 of them are easy to be found. They are (0,3), (6,3) and (6,6).
The last one is the common point of lines
x  4 y  12 and x  2 y  0. Therefore we setup
 x  4 y  12    (1)

 x  2 y  0    (2)
(1)  (2), we can get
6 y  12  y  2
Plug into (2),
x40 x  4
So the last coner point
is (4, 2)

17
Now we calculate value of z  x  10 y at those 4 corner points and
draw the following table

Corner
Value z = x + 10y
points
(0, 3)       0+3(10)=30
(6, 3)      6+10(3) = 36
6+10(6) = 66
(6, 6)
maximum
(4, 2)      4+10(2) = 24
minimum

The maximum value of z is 66 at point (6,6)
The minimum value of z is 24 at point (4,2)
18
Exercises
Find maximum and minimum values in the region for the functions
1. z  3 x  2 y                 2. z  0.4 x  0.75 y
(2, 8)
(0, 12)

(4, 6)                    (4, 8)
(1, 4)

(7, 3)

(4, 1)
(0, 0)               (8, 0)

19
3. Find maximum and minimum of z  4 x  y if they exist

(0, 10)

(2, 4)

(5, 2)
(15, 0)

20
4. Find maximum and minimum of z  3x  6 y in region
 x  y  10

5 x  2 y  20
 x  2y  4


21

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