Hi My question is:

A firm has production function f (x1, x2, x3, x4) = min {x1, x2} + min
{x3, x4}. This firm
faces competitive factor markets where the prices for the four factors
are w1 = $4, w2 = $8,
w3 = $5, and w4 = $3. The firm must use at least 20 units of factor 2.
The cost of producing
100 units in the cheapest possible way is
A. $700.
B. $300.
C. $880.
D. $1,300.
E. $1,040.

The way I attempted this question was:
100 = 20 + min{x3,x4)
Therefore they pick 80 of x4 as its cheaper so cost = 400
But this is wrong!
PLEASE HELP!

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Clarification of Question by irish_traveller-ga on 28 May 2006 09:30 PDT

anything I can do to clarify it?

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Clarification of Question by irish_traveller-ga on 28 May 2006 14:06 PDT

The answer is C but I have no idea why?!!

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Request for Question Clarification by livioflores-ga on 28 May 2006 14:40 PDT

I am working on this question. I will post an answer tonight, is this good for you?

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Clarification of Question by irish_traveller-ga on 28 May 2006 15:22 PDT

Hi! An answer asap would be great :)
But if you still need time thats no problem you can post it "tonight"
(im on a different timescale so I'd get it in the morning which is
fine)

Hi!! 

I got the answer! 

Start with this: 
100 = min(x1,20) + min(x3,x4) 

Since you want to minimize costs  you do not want to use more than 20
units of x1, then you have:
100 = 20 + min(x3,x4) 
then: 
min(x3,x4) = 80 
Again you want to minimize costs, so you will use the minimum amount
possible of x3 and x4, that is:
x3 = x4 = 80 

Now the cost of the production is: 
Cost = 20*4 + 20*8 + 80*5 + 80*3 =
     = $880 

So yes, the answer is C. 

I hope this helps you. Feel free to use the clarification feature if you need it. 

Regards, 
livioflores-ga

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Request for Answer Clarification by irish_traveller-ga on 28 May 2006 18:13 PDT

as you want: min(x3,x4) = 80 
how do u know x3 = x4 = 80?
I hate production functions! I have another two similiar question but
I'll go broke if I keep submitting them (I'm a student), can I ask
them here?

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Request for Answer Clarification by irish_traveller-ga on 28 May 2006 18:15 PDT

Wait why did u use 20 of x1 too?
AHHHHH!

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Clarification of Answer by livioflores-ga on 28 May 2006 19:14 PDT

Hi!!


You asked "how do u know x3 = x4 = 80?":

Since 100 = 20 + min(x3,x4) 
then: 
min(x3,x4) = 80 
This means that both factors are:
x3 >= 80 and x4 >= 80,
If you want to minimize the costs you must use the minimum amount
possible in both cases, that is 80 for both factors.

Hope this helps you!!

Regards,
livioflores-ga

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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