Question 1166148
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            I will present you another approach.



<pre>
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.


<U>ANSWER</U>.  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.
</pre>

Solved.


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