Question

A coffee manufacturer is interested in blending three types of coffee beans into 10,000
pounds of a final coffee blend. The three component beans cost $2.40, $2.60, and $2.00
per pound, respectively. The manufacturer wants to blend the 10,009 pounds at a total
cost of $21,000. In blending the coffee, one restriction is that the amounts used of
component beans 1 and 2 be the same. Determine whether there is a combination of the
three types of beans which will lead to a final blend of 19,000 pounds costing $21,000
and satisfying the blending restriction.​

Answer 1

Bean 1 = 1000 pound , Bean 2 = 1000 pound  , Bean 3 = 8000 Pound  for coffee manufacturer for 10000 pounds in $21000

Step-by-step explanation:

beans 1 and 2 be the same  & total = 10000 pound

Bean 1  =  B  Pound

Bean 2 = B   Pound

Bean 3  =  10000 - 2B   Pound

Total Cost = 2.4B  + 2.6B  + 2(10000 - 2B)  

Total Cost = $ 21000

=> 5B  + 20000 - 4B = 21000

=> B = 1000

Bean 1 = 1000 pound

Bean 2 = 1000 pound

Bean 3 = 8000 Pound

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Answer 2

Answer:

1000 pounds of bean 1

1000 pounds of bean 2

8000 pound of bean 3

Step-by-step explanation:

We know that  the total amount required is 10000 pounds, the costs of the three types of coffee beans and that quantity of bean 1 and bean 2 must be equal.

Let the quantity of Bean 1 be denoted as $x$,

Then the quantity of Bean 2 is $x$,

And that of Bean 3 is given as

$10000-x-x=10000-2x$

From the cost function we have the following equation;

$2.4x+2.6x+2(10000-2x)=21000\\5x+20000-4x=21000\\5x-4x=21000-20000\\x=1000$

Therefore the combination required is;

1000 pounds of bean 1

1000 pounds of bean 2

(10000-2*1000) Pounds of bean 3

ie 8000 pound of bean 3