**
Clarification of Answer by
livioflores-ga**
on
28 May 2006 20:01 PDT
Hi!!
Here is the answer to your second request of clarification:
You know that 100 = min(x1,20) + min(x3,x4);
Since you are trying to minimize costs it is clear that x1=<20; if not
is x1>20 but min{x1,20} is still equal to 20 and this force you to
continue using 80 units of x3 and x4; but in this situation you will
spend more that 20 units of x1 when you can use 20 or less.
In the above paragraph we ruled out the option x1 > 20 ; so now
consider that x1 =< 20:
First of all take into account the conclusion taken in the first
clarification, once the number min{x1,20} is found, this number will
determine the value of min{x3,x4}; and whatever this number is, it
will be:
x3 = x4 = 100 - min{x1,20} .
Now consider the cost function:
Cost = x1*4 + 20*8 + x3*5 + x4*3 =
= x1*4 + 20*8 + [100 - min{x1,20}]*5 + [100 - min{x1,20}]*3 =
= x1*4 + 160 + [100 - min{x1,20}]*(5+3) =
= x1*4 + 160 + [100 - min{x1,20}]*8 ;
Since we are considering the case x1 < 20 it is:
min{x1,20} = x1 ,
then:
Cost = x1*4 + 160 + [100-x1]*8 =
= 160 + x1*4 + 800 - x1*8 =
= 960 - x1*4 , with x1 =< 20
Now it is clear that the cost is minimized when x1 takes its highest
possible value, due the restriction this value is 20, then the minimum
cost is:
Cost = 960 - 20*4 = 960 - 80 = $880
I think that this is the better way to solve this problem, so putting
all things in order the more appropiated answer is:
CORRECTED ANSWER:
100 = min(x1,20) + min(x3,x4)
Then:
min(x3,x4) = 100 - min(x1,20) = M
Given M, this means that both factors are:
x3 >= M and x4 >= M,
If you want to minimize the costs you must use the minimum amount
possible in both cases, that is M for both factors.
You know that x2=20, and you want to minimize costs, so you do not
want to use more than 20 units of x1, then you have that x1 =< 20 .
If not is x1 > 20 but min{x1,20} is still equal to 20 and this force
you to continue using 80 units of x3 and x4; but in this situation you
will spend on more than 20 units of x1 when you can use only 20 or
less.
Now consider the cost function:
Cost = x1*4 + 20*8 + x3*5 + x4*3 =
= x1*4 + 20*8 + [100 - min{x1,20}]*5 + [100 - min{x1,20}]*3 =
= x1*4 + 160 + [100 - min{x1,20}]*(5+3) =
= x1*4 + 160 + [100 - min{x1,20}]*8 ;
Since we are considering the case x1 =< 20 it is:
min{x1,20} = x1 ,
then:
Cost = x1*4 + 160 + [100-x1]*8 =
= 160 + x1*4 + 800 - x1*8 =
= 960 - x1*4 , with x1 =< 20
Now it is clear that the cost is minimized when x1 takes its highest
possible value, due the restriction this value is 20, then the minimum
cost is:
Cost = 960 - 20*4 = 960 - 80 = $880
Hope this helps you.
Best regards,
livioflores-ga