SOLUTION: a chicken farmer can buy a special food mix A at 20c per kg and special food mix B at 40c per kg each kg of mix A contains 3000 units of nutrient N1 and 1000 units nutrient of N2

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Question 1166148: a chicken farmer can buy a special food mix A at 20c per kg and special food mix B at 40c per kg each kg of mix A contains 3000 units of nutrient N1 and 1000 units nutrient of N2 each kg of mix B contains 4000 units of nutrient N1 and 4000 units of nutrients N2 if the minimum daily requirements for the chickens collectively are are 36000 units of nutrient N1 and 20000 units of nutrient N2 how many pounds of each food mix should be used each day to minimize daily food costs while meeting (or exceeding) the minimum daily requirements? what is the minimum daily cost?
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


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Where is the number of kilos of mix A and is the number of kilos of mix B.

Subject to:










This is a simple two constraint problem. You can graph it or use the LP Simplex method or the Solver Add-In in Excel. The set up is the hard work. I did that for you. Now you need to do the scut work.


John

My calculator said it, I believe it, that settles it


I > Ø

Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.

            I will present you another approach. 


From the condition, you have this system of 2 equations in 2 unknowns


    3000A + 4000B = 36000   (1)    (nutrient A)

    1000A + 4000B = 20000   (2)    (nutrient B)


To solve the system, subtract equation (2) from equation (1).  You will get


    2000A        = 36000 - 20000 = 16000

        A                        = 16000/2000 = 8.


Then from equation (2),


    1000*8 + 4000B = 20000

    4000B          = 20000 - 1000*8 = 12000

        B                           = 12000/4000 = 3.


ANSWER.  A = 8,  B = 3  satisfies to BOTH requirements to provide the necessary minimum nutrients and to minimize cost.


         The minimum cost is thus  20 c/kilo * 8 kilograms + 40 c/kilo * 3 kilograms = 160 c + 120 c = 280 c = $2.80.

Solved.

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In the problem, the question is about  "how many pounds",  but  I think that it is a  mistake/typo,
and in reasonably posed problem the question must ask about  kilograms.