\r\n" );
document.write( "Let x = number of grams of soybean meal\r\n" );
document.write( "Let y = number of grams of meat byproducts\r\n" );
document.write( "Let z = number of grams of grain\r\n" );
document.write( "We are to minimize:\r\n" );
document.write( "
\r\n" );
document.write( "subject to the constraints:\r\n" );
document.write( "
\r\n" );
document.write( "We use the simplex method.\r\n" );
document.write( "We form the initial matrix\r\n" );
document.write( "
\r\n" );
document.write( "Since this is a minimizing problem, we form the dual\r\n" );
document.write( "maximizing problem. We form the transpose of the\r\n" );
document.write( "above matrix:\r\n" );
document.write( "
\r\n" );
document.write( "The dual maximizing problem:\r\n" );
document.write( "Maximize:\r\n" );
document.write( "
\r\n" );
document.write( "subject to the constraints:\r\n" );
document.write( "
\r\n" );
document.write( "
\r\n" );
document.write( "\r\n" );
document.write( "We form the initial tableau:\r\n" );
document.write( "\r\n" );
document.write( "
\r\n" );
document.write( "\r\n" );
document.write( "The most negative number on the bottom row is -66.\r\n" );
document.write( "It is in column 2, so we call column 2 \"the pivot column\".\r\n" );
document.write( "We divide each positive number above the pivot element INTO the number\r\n" );
document.write( "at the right to see which gives the smallest positive answer.\r\n" );
document.write( "\r\n" );
document.write( " 1.4 3 1.1\r\n" );
document.write( "5)7.0 3)9 10)11.0\r\n" );
document.write( "\r\n" );
document.write( "We get the smallest value when we divide 10 into the number at the far\r\n" );
document.write( "right, so 10 is the pivot element.\r\n" );
document.write( "\r\n" );
document.write( "Now we make the pivot element become 1, by dividing the pivot row\r\n" );
document.write( "through by 10. Then we make 0's elsewhere in the pivot column, using\r\n" );
document.write( "the pivot row. Making an element of a column become 1 and all the other\r\n" );
document.write( "elements in the column become 0 is called \"pivoting\" on the element\r\n" );
document.write( "that we caused to become 1.\r\n" );
document.write( "\r\n" );
document.write( "
\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now the only negative number on the bottom row is -27. It is in \r\n" );
document.write( "column 1, so now column 1 is the pivot column. We divide each \r\n" );
document.write( "positive number above the pivot element INTO the number\r\n" );
document.write( "at the right to see which gives the smallest positive answer.\r\n" );
document.write( "\r\n" );
document.write( " 1.9 2.2\r\n" );
document.write( "3)5.7 0.5)1.1\r\n" );
document.write( "\r\n" );
document.write( "We get the smallest value when we divide 3 into the number at the far\r\n" );
document.write( "right, so 3 is the pivot element.\r\n" );
document.write( "\r\n" );
document.write( "Now we make the pivot element become 1, by dividing the pivot row\r\n" );
document.write( "through by 3. Then we use it to make 0's elsewhere in the pivot column,\r\n" );
document.write( "using the pivot row.\r\n" );
document.write( "\r\n" );
document.write( "
\r\n" );
document.write( "\r\n" );
document.write( "There are no more negative numbers on the bottom row, so\r\n" );
document.write( "we have reached the final tableau:\r\n" );
document.write( " \r\n" );
document.write( "So the minimum cost is 123.9, or $123.90 and this occurs when\r\n" );
document.write( "x1 = 0, the number at the bottom of the s1 column. x2 = 9, \r\n" );
document.write( "the number at the bottom of the s2 column. x3 = 3.9\r\n" );
document.write( "the number at the bottom of the s3 column.\r\n" );
document.write( "\r\n" );
document.write( "Note this is the same answer that Ikleyn got using a different technique.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
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document.write( "