document.write( "Question 317370: The initil tableau of a linear programming problem is given below. Use the simplex method to solve.\r
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Algebra.Com's Answer #227276 by Edwin McCravy(20055) $\"\"$ $\"About$
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document.write( "The original problem was\r\n" );
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document.write( "Maximize  subject to\r\n" );
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document.write( "Then slack variables were , ,  were\r\n" );
document.write( "introduced to turn the inequalities into equations and the\r\n" );
document.write( "objective function equation rearranged at the bottom with 0 on\r\n" );
document.write( "the right side:\r\n" );
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document.write( "Then it was written as\r\n" );
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document.write( "And this system of equations was written as an augmented matrix:\r\n" );
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document.write( "The most negative number (indicator) on the bottom row is -3.  \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 -3 INTO the\r\n" );
document.write( "element at the far right of its row:\r\n" );
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document.write( "   5       15\r\n" );
document.write( "2)10     1)15\r\n" );
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document.write( "The smallest of 5 and 15 is 5, which was gotten using the elements\r\n" );
document.write( "of row 1, so row 1 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.\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 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 in the 2nd row 1st column a 0 by multiplying the 1st\r\n" );
document.write( "row temporarily by -1 and adding it to row 2.\r\n" );
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document.write( "We make the -3 in the 3rd row 1st column a 0 by multiplying the 1st\r\n" );
document.write( "row temporarily by 3 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 z, the letter to maximize:\r\n" );
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document.write( "Since , , and  are non-negative,\r\n" );
document.write( "the maximum value z can take on is 15, when , , \r\n" );
document.write( "and  are all 0, so we substitute 0 for , \r\n" );
document.write( ", 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 z 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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