document.write( "Question 172970: Set up the simplex matrix used to solve this linear programming problem. all variables are nonnegative.
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Algebra.Com's Answer #129794 by Edwin McCravy(20060) $\"\"$ $\"About$

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Set up the simplex matrix used to solve this linear programming problem. all variables are nonnegative.
\n" ); document.write( "Maximize $\"f=3x%2B5y%2B11z\"$ subject to
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document.write( "Introduce slack variables , , and ,\r\n" );
document.write( "to \"take up the slack\" between the left sides\r\n" );
document.write( "and the right sides of the first three and the\r\n" );
document.write( "objective function, the one to maximize, at the bottom: \r\n" );
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document.write( "Rearrange the terms in the bottom equation\r\n" );
document.write( "and put in 0 and 1 coefficients of all the \r\n" );
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document.write( "Then we set up the matrix with petitions as follows:\r\n" );
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document.write( "  \r\n" );
document.write( "That matrix is called \"the initial tableau\".\r\n" );
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document.write( "That's all you were asked for, but I thought\r\n" );
document.write( "I'd show you how to solve it.\r\n" );
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document.write( "The numbers on the bottom left, \r\n" );
document.write( "all but the last one, are called \"indicators\",\r\n" );
document.write( "Pick the indicator that is the most negative.  \r\n" );
document.write( "That is, the . It is in column 3, so \r\n" );
document.write( "we call column 3 the \"pivot column\".\r\n" );
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document.write( "Next divide each of the three numbers above it into the number\r\n" );
document.write( "at the far right of the row it's in. That is we do these\r\n" );
document.write( "three divisions:\r\n" );
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document.write( "  15    48        48\r\n" );
document.write( "4)60, 1)48, and 1)48 \r\n" );
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document.write( "The smallest of these is 15, obtained from row 1,\r\n" );
document.write( "So we call row 1 the \"pivot row\".\r\n" );
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document.write( "And we call the number in both the pivot row and the\r\n" );
document.write( "pivot column the \"pivot element\", which is 4.\r\n" );
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document.write( "Now we divide the pivot row by that number, 4, to make \r\n" );
document.write( "the pivot element become 1.\r\n" );
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document.write( "Now we use the pivot row to get 0's everywhere\r\n" );
document.write( "else in the pivot column.\r\n" );
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document.write( "Then we get a 0 where the 1 is in row 2 by multiplying\r\n" );
document.write( "-1 times the pivot row to row 2. \r\n" );
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document.write( "Then we get a 0 where the 1 is in row 2 by multiplying\r\n" );
document.write( "-1 times the pivot row to row 3.\r\n" );
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document.write( "Then we get a 0 where the -11 is in the bottom row by \r\n" );
document.write( "multiplying 11 times the pivot row to the bottom row.\r\n" );
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document.write( "Now we have:\r\n" );
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document.write( "Usually some of the indicators on the bottom\r\n" );
document.write( "row will still be negative, and we would have \r\n" );
document.write( "to repeat the operation until there were no\r\n" );
document.write( "negative indictors on the bottom row.  However \r\n" );
document.write( "this particular problem became solved with just \r\n" );
document.write( "one operation, so we convert this into a system \r\n" );
document.write( "of equations:\r\n" );
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document.write( "Now look at the bottom equation:\r\n" );
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document.write( "Solve for \r\n" );
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document.write( "Since none of the variables can be negative,\r\n" );
document.write( "f will be the largest value it can possibly be\r\n" );
document.write( "when nothing is subtracted from the 165, so \r\n" );
document.write( "that maximum is reached when ,, and  are \r\n" );
document.write( "all taken to be . So the system now becomes\r\n" );
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document.write( "or\r\n" );
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document.write( "So  has a maximum value of  when\r\n" );
document.write( ", , and \r\n" );
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document.write( "Edwin

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