Question 1129285
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A farmer can buy two types of plant food, mix A and mix B. 
Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of​ nitrogen, and 5 pounds of potash. 
Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash. 
The minimum monthly requirements are 400 pounds of phosphoric acid, 990 pounds of nitrogen, and 210 pounds of potash. 
If mix A costs $20 per cubic yard and mix B costs $40 per cubic yard, how many cubic yards of each mix should the farmer blend 
to meet the minimum monthly requirements at a minimum cost? What is this cost?
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<pre>
Let X be the amount of the mix A (in pounds), and Y be the amount of the mix B.


The constraints are

20X + 10Y >= 400     (1)   (phosphoric acid)
30X + 30Y >= 990     (2)   (nitrogen)
 5X + 10Y >= 210     (3)   (potash)

X >= 0,  Y >= 10.


The objective function to minimize is  F(X,Y) = 20X + 40Y.


In equivalent (and simplified) form the constraints are

2X + 1Y >= 40        (1') 
 X +  Y >= 33        (2') 
 X + 2Y >= 42        (3') 


The feasibility area is the infinite area in the first quadrant which lies ABOVE each of the following straight lines : 

    - red   (which represents the constraint (1'));
    - green (which represents the constraint (2')),  and
    - blue  (which represents the constraint (3')).


    {{{graph( 330, 330, -4, 50, -5, 50,
          40-2x, 33-x, (42-x)/2
)}}}


Plot  2X + 1Y = 40 (constr. (1'), red);   X +  Y = 33 (constr. (2'), green)  and   X + 2Y = 42 (constr. (3'), blue)


The four corner points of the feasibility area are

    P1 = (0,40)   (red line y-intercept);
    P2 = (7,26)   (red and green lines intersection);
    P3 = (24,9)   (green and blue lines intersection), and
    P4 = (42,0)   (blue line x-intercept).


According to the Linear Programming method, we should calculate and compare the values of the objective function at these 4 points.

These values are

    P1:  F(0,40) = 20*0 + 40*40 = 1600;
    P2:  F(7,26) = 20*7 + 40*26 = 1180;
    P3:  F(24,9) = 20*24 + 40*9 =  840;
    P4:  F(42,0) = 20*42 + 40*0 =  840.


Comparing these numbers, you see that any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements.


<U>ANSWER</U>.  Any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements.  The minimal price value is  $840.
</pre>

Solved.