Solution.
May be the condition of the task next:
z=x1+x2−x3−max,4x1+x2+x3=4,3x1+2x2−x3=6,x1,x2,x3>=0.
Introduce new variables x4,x5:
4x1+x2+x3+x4=4,3x1+2x2−x3+x5=6,
then z=−Mx4−Mx5==−M(4−4x1−x2−x3)−M(6−3x1−2x2+x3)==7Mx1+Mx2−10M−max
Introduce new variable x_0:
x0=7x1+3x2.
We will have:
x0=−10+7x1+3x2,
x4=4−4x1−x2−x3,x5=6−3x1−2x2+x3.
This plan is not optimal because there are positive elements in the expression for x0 . Select a new variable x1, then
x1=1−41x2−41x3−41x4.
From here
x0=−3+45x2−47x3−47x4,
x5=3−45x2+47x3+43x4.
The expression for x0 has positive elements, the plan is not optimal. The largest coefficient is near x2 , then choose x2 as a new variable.
x2=512+57x3+53x4−54x5,x0=0−x4−x5,x1=52−53x3−52x4+51x5.
The expression for x0
has not positive elements, the plan is optimal.
We will deduce variables x4,x5 from basis:
x0=514−51x3,
x1=52−53x3,
x2=512+57x3.
Optimal plan:
x1=52,x2=252,x3=0.
So, max z(x)=252.
Answer. max z(x)=252.
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