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Consider the linear programming Problem: Maximize $z= 2x_1 +3x_2 +x_3$ such that

$4x_1+3x_2 + x_3=6$

$x_1 + 2x_2 + 5x_3 \ge 4$

$x_1, x_2 , x_3 \ge 0$

Write down the objective function of the dual problem

My attempt : From $4x_1+3x_2 + x_3=6$ - $(1)$

$x_1 + 2x_2 + 5x_3 \ge 4$ -$(2)$

Multiply $(2)$ with $4$ we have $4x_1 +8x_2 + 20x_3 =16$

Now $4x_1 +8x_2 + 20x_3 =16$ subtract $4x_1+3x_2 + x_3=6$ we have $6x_2 +19x_3 =10$ After that im not able proceed further

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jasmine
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1 Answers1

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From the primal problem right hand sides, the objective of the dual problem is to minimize $6y_1+4y_2$.

RobPratt
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  • How u got $ 6y_1+4y_2$? Can u elaborate more @Rob – jasmine Feb 27 '20 at 04:50
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    The primal is maximization, so the dual is minimization. The primal has two constraints, so the dual has two variables, say $y_1$ and $y_2$, and their objective coefficients are the primal rhs values. Because the first primal constraint is equality, $y_1$ is free. Because the second constraint is $\ge$ in a maximization, $y_2\le 0$. – RobPratt Feb 27 '20 at 04:56