document.write( "Question 251640: A company makes three products, A, B, and C. There are 500 pounds of raw material available. Each unit of product A requires 2 pounds of raw material, each unit of product B requires 2 pounds, and each unit of product C requires 3 pounds. The assembly line has 1,000 hours of operation available. Each unit of product A requires 4 hours, while each unit of products B and C requires 5 hours. The company realizes a profit of $500 for each unit of product A, $600 for each unit of product B, and $1,000 for each unit of product C. How many units of each of the three products should the company make to maximize profits?\r
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Algebra.Com's Answer #185345 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
A company makes three products, A, B, and C. There are 500 pounds of raw material available. Each unit of product A requires 2 pounds of raw material, each unit of product B requires 2 pounds, and each unit of product C requires 3 pounds. The assembly line has 1,000 hours of operation available. Each unit of product A requires 4 hours, while each unit of products B and C requires 5 hours. The company realizes a profit of $500 for each unit of product A, $600 for each unit of product B, and $1,000 for each unit of product C. How many units of each of the three products should the company make to maximize profits?
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document.write( "Let x = the number of units of A\r\n" );
document.write( "Let y = the number of units of B\r\n" );
document.write( "Let z = the number of units of C\r\n" );
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document.write( "Let's make an inequality for the raw materials only:\r\n" );
document.write( " \r\n" );
document.write( "\"2x+%2B+2y++%2B+3y+%3C=+500\"\r\n" );
document.write( " \r\n" );
document.write( "Let's make an inequality for the hours of operation only:\r\n" );
document.write( " \r\n" );
document.write( "\"4x+%2B+5y+%2B+5z+%3C=+1000\"\r\n" );
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document.write( "Let's make an equation for the profit P.\r\n" );
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document.write( "\"P+=+500x+%2B+600y+%2B+1000z\"\r\n" );
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document.write( "So the problem becomes this:\r\n" );
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document.write( "Maximize\r\n" );
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document.write( "\"P+=+500x+%2B+600y+%2B+1000z\"\r\n" );
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document.write( "subject to the constraints:\r\n" );
document.write( " \r\n" );
document.write( "\"2x+%2B+2y++%2B+3y+%3C=+500\"\r\n" );
document.write( "\"4x+%2B+5y+%2B+5z+%3C=+1000\"\r\n" );
document.write( "\"P+=+500x+%2B+600y+%2B+1000z\"\r\n" );
document.write( " \r\n" );
document.write( "and of course \"x%3E=0\", \"y%3E=0\" and \"z%3E=0\"\r\n" );
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document.write( "we introduce non-negative slack variables \"s%5B1%5D\" \r\n" );
document.write( "and \"s%5B2%5D\" to make equations out of the two\r\n" );
document.write( "inequalities, and rewrite the equation for P getting\r\n" );
document.write( "0 on the right by subtracting the whol right side from\r\n" );
document.write( "both sides:\r\n" );
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document.write( "2x + 2y + 3z + s1 = 500\r\n" );
document.write( "4x + 5y + 5z + s2 = 1000\r\n" );
document.write( "-500x - 600y - 1000z + P = 0\r\n" );
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document.write( "Now we write it so that all the variables appear in all\r\n" );
document.write( "the equations with a coefficient.  When an equation does \r\n" );
document.write( "not contain a particular variable, it is placed in that \r\n" );
document.write( "equation with a 0 coefficient.  Also we will put in any\r\n" );
document.write( "understood coefficients of 1. We will write the equations\r\n" );
document.write( "so that like letters line up vertically:\r\n" );
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document.write( "    2x +   2y +    3z + 1s1 + 0s2 + 0P =  500\r\n" );
document.write( "    4x +   2y +    5z + 0s1 + 1s2 + 0P = 1000\r\n" );
document.write( " -500x - 600y - 1000z + 0s1 + 0s2 + 1P =    0\r\n" );
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document.write( "Put that in partitioned matrix form\r\n" );
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document.write( "    x      y       z    s1    s2    P\r\n" );
document.write( " ----------------------------------------------\r\n" );
document.write( "|   2      2       3  | 1     0  |  0  |  500 |\r\n" );
document.write( "|   4      2       5  | 0     1  |  0  | 1000 |\r\n" );
document.write( "|---------------------------------------------|\r\n" );
document.write( "|-500   -600   -1000  | 0     0  |  1  |    0 |\r\n" );
document.write( " ----------------------------------------------\r\n" );
document.write( " \r\n" );
document.write( "The indicators are the numbers on the bottom row.\r\n" );
document.write( "We want to get rid of all the negative indicators\r\n" );
document.write( "We start with the most negative one, the -1000.\r\n" );
document.write( "That's the pivot column.  I'll color it red and\r\n" );
document.write( "the two top numbers in the far right blue:\r\n" );
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document.write( "    x      y       z    s1    s2    P\r\n" );
document.write( " ----------------------------------------------\r\n" );
document.write( "|   2      2       3  | 1     0  |  0  |  500 |\r\n" );
document.write( "|   4      2       5  | 0     1  |  0  | 1000 |\r\n" );
document.write( "|---------------------------------------------|\r\n" );
document.write( "|-500   -600   -1000  | 0     0  |  1  |    0 |\r\n" );
document.write( " ---------------------------------------------- \r\n" );
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document.write( "Divide each of those blue numbers by the corresponding red number.\r\n" );
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document.write( "500÷3 = 166 2/3 and 1000÷5=200\r\n" );
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document.write( "166 2/3 is the smallest so the pivot element is 5. We make it 1\r\n" );
document.write( "by multiplying row 1 by 1/3 and then we use that row to make \r\n" );
document.write( "all the the other numbers in the red column zero.\r\n" );
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document.write( "    x      y   z       s1  s2     P\r\n" );
document.write( " ------------------------------------------------\r\n" );
document.write( "|  2/3    2/3  1  |   1/3   0  |  0  |    500/3 |\r\n" );
document.write( "|  2/3   -4/3  0  |  -5/3   1  |  0  |    500/3 |\r\n" );
document.write( "|-----------------------------------------------|\r\n" );
document.write( "|500/3  200/3  0  |1000/3   0  |  1  | 500000/3 |\r\n" );
document.write( " ------------------------------------------------\r\n" );
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document.write( "Now there are no negative indicators on the bottom\r\n" );
document.write( "row, so that is the final tableau.  If there had been\r\n" );
document.write( "any negative numbers on the bottom row we would have\r\n" );
document.write( "had to go through the process again.\r\n" );
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document.write( "next we write the equation form:\r\n" );
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document.write( "  \r\n" );
document.write( "  2/3x +   2/3y + z  +  1/3s1           = 500/3 \r\n" );
document.write( "  2/3x -   4/3y       - 5/3s1 + 1s2     = 500/3 \r\n" );
document.write( "500/3x + 200/3y    + 1000/3s1      + P = 500000/3\r\n" );
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document.write( "Solve the bottom equation for P:\r\n" );
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document.write( "p = 500000/3 - (500/3)x - (200/3)y - (1000/3)s1\r\n" );
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document.write( "So the maximum value of P is when we don't subtract\r\n" );
document.write( "anything from the 500000/3, which is when \r\n" );
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document.write( "x=0, y=0, and s1=0\r\n" );
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document.write( "Substituting in the first equation:\r\n" );
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document.write( "2/3(0) + 2/3(0) + z  +  1/3(0) = 500/3\r\n" );
document.write( "                             z = 500/3\r\n" );
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document.write( "Substituting in the second equation:\r\n" );
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document.write( "  2/3(0) - 4/3(0)  - 5/3(0) + 1s2  = 500/3  \r\n" );
document.write( "                               s2  = 500/3\r\n" );
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document.write( "So the maximum profit of $500000/3 or $166666.67 is obtained when\r\n" );
document.write( "no A's or B's are made and 500/3 or \"166%262%2F3\" C's are made.\r\n" );
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document.write( "Since you have to make whole products, you can only make 166 C's\r\n" );
document.write( "and that profit is $166000.  \r\n" );
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document.write( "Your teacher probably expects that answer. But if you want to get\r\n" );
document.write( "precise, you go further:\r\n" );
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document.write( "If you only make 166 C's, then you have used up only 498 pounds of\r\n" );
document.write( "raw material, and you have 2 pounds of raw materials left over.\r\n" );
document.write( "That's enough to raw material to make either 1 A or 1 B.  We would \r\n" );
document.write( "choose to make 1 B since the profit is more, $600.\r\n" );
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document.write( "And you've used up only 830 of the 1000 hours of operation,\r\n" );
document.write( "so you have plenty of operation hours, in fact, 170 left over.\r\n" );
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document.write( "So the exact answer is to make 166 C's and 1 B for a maximum\r\n" );
document.write( "profit of $166000 + $600 = $166600.\r\n" );
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document.write( "Edwin
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