SOLUTION: Find out the maximum and Minimum value of (a) |z-2|+|z-3| (b) |2z-1|+|3z-2|

We consider 4 potential cases

 Case 1:  both expressions within absolute value bars 
are non-negative:

  

That's the same as 

So |z-2|+|z-3| becomes  

(z-2)+(z-3) = z-2+z-3 = 2z-5

We build up 2z-5 from 







So the minimum value is 1
and there is no maximum value.

Case 2:  the first expressions within absolute value bars 
is non-negative and the second expression within absolute
value bars is non-positive



That's the same as .

 So  |z-2|+|z-3| becomes 

(z-2)-(z-3) = z-2-z+3 = 1

So the minimum (and maximum) value for case 2 is 1.

Case 3:  the first expressions within absolute value bars 
is non-positive and the second expression within absolute
value bars is non-negative:



That is impossible, so we rule out case 3

Case 4:  both expressions within absolute value bars 
are non-positive:

 

That's the same as 

So  |z-2|+|z-3| becomes  

 -(z-2)-(z-3) = -z+2-z+3 = -2z+5

We build up -2z+5 from 


  <--inequality reverses




So for case 4, the minimum value is 1 and there is no maximum value.

Therefore for all the cases the minimum value is 1
and there is no maximum value. 

--------------------------------------

 
Case 1:  both expressions within absolute value bars 
are non-negative:

 

 

That's the same as 

So |2z-1|+|3z-2| becomes  

(2z-1)+(3z-2) = 2z-1+3z-2 = 5z-3

We build up 5z-3 from 








That has a minimum value of 7/3
and no maximum value.
 
Case 2:  the first expressions within absolute value bars 
is non-negative and the second expression within absolute
value bars is non-positive


z>=1/2,  and     x<=2/3)}}}

 
That's the same as .

So  |2z-1|+|3z-2| becomes 
    (2z-1)-(3z-2) = 2z-1-3z+2 = -z+1 
 
We build up -z+1 from .

 
   <---the inequality symbol reverses 



So the minimum value for case 2 is 1/3
And the maximum value for case 2 is 1/2

Case 3:  the first expressions within absolute value bars 
is non-positive and the second expression within absolute
value bars is non-negative:

 
z<=1/2,  and,    z>=2/3}}}
 
That is impossible, so we rule out case 3.

Case 4:  both expressions within absolute value bars 
are non-positive:

 

That's the same as 

So |2z-1|+|3z-2| becomes  

-(2z-1)-(3z-2) = -2z+1-3z+2 = -5z+3

We build up -5z+3 from 


   <--inequality symbol reverses





That has a minimum value of 1/2
and no maximum value.

The minimum value from all the cases is 1/3 and
there is no maximum value.

Edwin