document.write( "Question 1206990: solve using simplex method max z=3x1+X2+3x3 for the following constraints 2x1+x2+x3≤2, x1+2x2+3x3≤5, 2x1+2x2+x3≤6 \n" ); document.write( "
Algebra.Com's Answer #844798 by mccravyedwin(407)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Maximize:\r\n" );
document.write( "\"z=3x%5B1%5D%2Bx%5B2%5D%2B3x%5B3%5D\"\r\n" );
document.write( "subject to the constraints:\r\n" );
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document.write( "We add non-negative slack variables to \"take up the slack\" in the inequalities\r\n" );
document.write( "to change them into equations. We rearrange the equation of the variable z to\r\n" );
document.write( "maximize and put it at the bottom:\r\n" );
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document.write( "We form the initial tableau:\r\n" );
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document.write( "The most negative number on the bottom row is -3. We find it twice at the bottom\r\n" );
document.write( "of columns 1 and 3. So we could either pick column 1 or column 3.\r\n" );
document.write( "We'll pick column 1, so we call column 1 \"the pivot column\".\r\n" );
document.write( "We divide each positive number in the pivot column INTO the number\r\n" );
document.write( "at the far right to see which gives the smallest positive answer.\r\n" );
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document.write( "  1         5         3\r\n" );
document.write( "2)2       1)5       2)6\r\n" );
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document.write( "We get the smallest value when we divide 2 into the number at the far\r\n" );
document.write( "right, so 2 in column 1 row 1 is the pivot element.  Its row, the 1st row,\r\n" );
document.write( "is the pivot element'\r\n" );
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document.write( "Now we make the pivot element become 1, by dividing the pivot row\r\n" );
document.write( "through by 2.  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.  So here is the 2nd tableau:\r\n" );
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document.write( "It is not the final tableau, because there is still a negative number\r\n" );
document.write( "on the bottom row.\r\n" );
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document.write( "The only negative number on the bottom row is -1.5. It is in \r\n" );
document.write( "column 3, so now column 3 is the pivot column. We divide each \r\n" );
document.write( "positive number in the pivot column INTO the number\r\n" );
document.write( "at the far right to see which gives the smallest positive answer.\r\n" );
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document.write( "    2          1.6\r\n" );
document.write( "0.5)1       2.5)4\r\n" );
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document.write( "We get the smallest value when we divide 2.5 into the number at the far\r\n" );
document.write( "right, so 2.5 is the pivot element.\r\n" );
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document.write( "Now we make the pivot element become 1, by dividing the pivot row\r\n" );
document.write( "through by 2.5.  Then we use it to make 0's elsewhere in the pivot column,\r\n" );
document.write( "which is column 3, using the pivot element.\r\n" );
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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" );
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document.write( "Now we turn the final tableau back into a system of equations:\r\n" );
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document.write( "We solve the last equation for z:\r\n" );
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document.write( "\"z=5.4-1.4x%5B2%5D-1.2s%5B1%5D-0.6s%5B2%5D\"\r\n" );
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document.write( "Now we see that the maximum value of z will be 5.4 if we don't\r\n" );
document.write( "subtract anything from it, so we choose x2=0, s1=0, s2=0.\r\n" );
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document.write( "Then the system above becomes:\r\n" );
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document.write( "\"system%28x%5B1%5D=0.2%2Cx%5B3%5D=1.6%2Cs%5B3%5D=4%2Cz=5.4%29\"\r\n" );
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document.write( "Or z has maximum value of 5.4 when x1=0.2, x2=0, and x3=1.6\r\n" );
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document.write( "Edwin
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