document.write( "Question 1092325: Primal Problem:
\n" ); document.write( "maximize profit P = 40x + 30y
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Algebra.Com's Answer #706974 by AnlytcPhil(1806) $\"\"$ $\"About$
You can put this solution on YOUR website!
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document.write( "Actually the DUAL Simplex method is for minimizing, not\r\n" );
document.write( "maximizing.  So we do not use the DUAL here but the\r\n" );
document.write( "regular Simplex method.\r\n" );
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document.write( "Maximize profit  subject to\r\n" );
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document.write( "Then non-negative slack variables ,  are\r\n" );
document.write( "introduced to turn the inequalities into equations and the\r\n" );
document.write( "objective function equation is rearranged at the bottom with\r\n" );
document.write( "0 on the right side:\r\n" );
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document.write( "Then it is written as\r\n" );
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document.write( "And this system of equations is written as an augmented matrix:\r\n" );
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document.write( "The most negative number (indicator) on the bottom row is -40.  \r\n" );
document.write( "It is in column 1, so column 1 is the PIVOT COLUMN\r\n" );
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document.write( "We now divide each of the positive numbers above the pivot element\r\n" );
document.write( "-40 INTO the element at the far right of its row:\r\n" );
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document.write( "    240             160\r\n" );
document.write( "  1)240           2)320\r\n" );
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document.write( "The smallest of 240 and 160 is 160, which was gotten using the elements\r\n" );
document.write( "of row 2, so row 2 is the PIVOT ROW.\r\n" );
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document.write( "So the element in the PIVOT ROW and the PIVOT COLUMN is the 2,\r\n" );
document.write( "which is called the PIVOT ELEMENT, boxed in red below.\r\n" );
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document.write( "We make the pivot element into a 1 by dividing the entire pivot row through\r\n" );
document.write( "by 2.\r\n" );
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document.write( "Now we make all the other numbers in the pivot column 0 by using this\r\n" );
document.write( "pivot row, multiplying it by whatever is necessary to multiply it by so\r\n" );
document.write( "that when we add it to the other row its first element will be 0.\r\n" );
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document.write( "We make the 1 in the 1st row 1st column a 0 by multiplying the 2nd\r\n" );
document.write( "row (the pivot row) temporarily by -1 and adding it to row 1.\r\n" );
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document.write( "We make the -40 in the bottom row 1st column a 0 by multiplying the 2nd\r\n" );
document.write( "row temporarily by 40 and adding it to row 3.\r\n" );
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document.write( "-----\r\n" );
document.write( "The most negative number (indicator) on the bottom row is -10.  \r\n" );
document.write( "It is in column 2, so now column 2 is the new PIVOT COLUMN\r\n" );
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document.write( "We now divide each of the positive numbers above the -10 INTO the\r\n" );
document.write( "element at the far right of its row.  So we divide 80 by 1/2,\r\n" );
document.write( "and 160 by 1/2. \r\n" );
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document.write( "80÷(1/2) = 80×(2/1) = 160\r\n" );
document.write( "10÷(1/2) = 160×(2/1) = 320\r\n" );
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document.write( "The smallest of 160 and 320 is 160, which was gotten using the elements\r\n" );
document.write( "of row 1, so row 1 is the new PIVOT ROW.\r\n" );
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document.write( "So the element in the PIVOT ROW and the PIVOT COLUMN is the 1/2,\r\n" );
document.write( "which is called the PIVOT ELEMENT, boxed in red below.\r\n" );
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document.write( "We make the pivot element into a 1 by dividing the entire pivot row through\r\n" );
document.write( "by 1/2, which is the same as multiplying it by 2:\r\n" );
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document.write( "Now we make all the other numbers in the pivot colomn 0 by using this\r\n" );
document.write( "pivot row, multipling it by whatever is necessary to multiply it by so\r\n" );
document.write( "that when we add it to the other row its first element will be 0.\r\n" );
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document.write( "We make the 1/2 in the 2nd row 2nd column a 0 by multiplying the 1st\r\n" );
document.write( "row (the pivot row) temporarily by -1/2 and adding it to row 2.\r\n" );
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document.write( "We make the -10 in the 3rd row 2nd column a 0 by multiplying the 1st\r\n" );
document.write( "row temporarily by 10 and adding it to row 3.\r\n" );
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document.write( "Now there are no more negative numbers on the bottom row. So we\r\n" );
document.write( "write the matrix as a system of equations:\r\n" );
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document.write( "Eliminate the zero terms and the 1 coefficients:\r\n" );
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document.write( "Now solve the bottom equation for P, the letter to maximize:\r\n" );
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document.write( "Since  and  are non-negative,\r\n" );
document.write( "the maximum value P can take on is 8000, when  and \r\n" );
document.write( " are both 0, so we substitute 0 for  and \r\n" );
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document.write( "and the system becomes:\r\n" );
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document.write( "or\r\n" );
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document.write( "So P reaches the maximum value of  when \r\n" );
document.write( " and .  And\r\n" );
document.write( "of course the slack variables  and .\r\n" );
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document.write( "Edwin
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