document.write( "Question 1129285: 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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Algebra.Com's Answer #745875 by ikleyn(52787)\"\" \"About 
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\n" ); document.write( "A farmer can buy two types of plant food, mix A and mix B.
\n" ); document.write( "Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of​ nitrogen, and 5 pounds of potash.
\n" ); document.write( "Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash.
\n" ); document.write( "The minimum monthly requirements are 400 pounds of phosphoric acid, 990 pounds of nitrogen, and 210 pounds of potash.
\n" ); document.write( "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
\n" ); document.write( "to meet the minimum monthly requirements at a minimum cost? What is this cost?
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document.write( "Let X be the amount of the mix A (in pounds), and Y be the amount of the mix B.\r\n" );
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document.write( "The constraints are\r\n" );
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document.write( "20X + 10Y >= 400     (1)   (phosphoric acid)\r\n" );
document.write( "30X + 30Y >= 990     (2)   (nitrogen)\r\n" );
document.write( " 5X + 10Y >= 210     (3)   (potash)\r\n" );
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document.write( "X >= 0,  Y >= 10.\r\n" );
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document.write( "The objective function to minimize is  F(X,Y) = 20X + 40Y.\r\n" );
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document.write( "In equivalent (and simplified) form the constraints are\r\n" );
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document.write( "2X + 1Y >= 40        (1') \r\n" );
document.write( " X +  Y >= 33        (2') \r\n" );
document.write( " X + 2Y >= 42        (3') \r\n" );
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document.write( "The feasibility area is the infinite area in the first quadrant which lies ABOVE each of the following straight lines : \r\n" );
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document.write( "    - red   (which represents the constraint (1'));\r\n" );
document.write( "    - green (which represents the constraint (2')),  and\r\n" );
document.write( "    - blue  (which represents the constraint (3')).\r\n" );
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document.write( "    \r\n" );
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document.write( "Plot  2X + 1Y = 40 (constr. (1'), red);   X +  Y = 33 (constr. (2'), green)  and   X + 2Y = 42 (constr. (3'), blue)\r\n" );
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document.write( "The four corner points of the feasibility area are\r\n" );
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document.write( "    P1 = (0,40)   (red line y-intercept);\r\n" );
document.write( "    P2 = (7,26)   (red and green lines intersection);\r\n" );
document.write( "    P3 = (24,9)   (green and blue lines intersection), and\r\n" );
document.write( "    P4 = (42,0)   (blue line x-intercept).\r\n" );
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document.write( "According to the Linear Programming method, we should calculate and compare the values of the objective function at these 4 points.\r\n" );
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document.write( "These values are\r\n" );
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document.write( "    P1:  F(0,40) = 20*0 + 40*40 = 1600;\r\n" );
document.write( "    P2:  F(7,26) = 20*7 + 40*26 = 1180;\r\n" );
document.write( "    P3:  F(24,9) = 20*24 + 40*9 =  840;\r\n" );
document.write( "    P4:  F(42,0) = 20*42 + 40*0 =  840.\r\n" );
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document.write( "Comparing these numbers, you see that any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements.\r\n" );
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document.write( "ANSWER.  Any value of (X,Y) that lies on the segment [P3,P4] satisfies the minimum requirements.  The minimal price value is  $840.\r\n" );
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