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Q: FIRM PRODUCTION FUNCTION COST MINIMIZATION ( Answered ,   0 Comments )
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 Subject: FIRM PRODUCTION FUNCTION COST MINIMIZATION Category: Business and Money > Economics Asked by: irish_traveller-ga List Price: \$15.00 Posted: 28 May 2006 07:21 PDT Expires: 27 Jun 2006 07:21 PDT Question ID: 733019
 ```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!``` Clarification of Question by irish_traveller-ga on 28 May 2006 09:30 PDT `anything I can do to clarify it?` Clarification of Question by irish_traveller-ga on 28 May 2006 14:06 PDT `The answer is C but I have no idea why?!!` 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?` 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``` 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?``` 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!``` 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``` 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```