State the dual problem for the linear programming problem.z = 7x₁ + 3x₂ + 8x3X₁ + x2 + x3 ≤6x₁ + x2≤37x₁ + x2 + 2x3 ≤ 15X₁20, X₂20, X3 20Maximizesubject to:withCState the dual problem.subject to:W =38withY₁20, Y₂20, Y3 20(Simplify your answers. Do not factor.)k

Answered: Maximize subject to: with z=7x₁ + 3x2 +…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

State the dual problem for the linear programming problem.
z = 7x₁ + 3x₂ + 8x3
X₁ + x2 + x3 ≤6
x₁ + x2
≤3
7x₁ + x2 + 2x3 ≤ 15
X₁20, X₂20, X3 20
Maximize
subject to:
with
C
State the dual problem.
subject to:
W =
3
8
with
Y₁20, Y₂20, Y3 20
(Simplify your answers. Do not factor.)
k

Expert Solution

Step 1: Finding the dual of the primal problem

Given primal problem is

Maximize $Z = 7 x_{1} + 3 x_{2} + 8 x_{3}$

subjected to

$x_{1} + x_{2} + x_{3} \leq 6 x_{1} + x_{2} \leq 3 7 x_{1} + x_{2} + 2 x_{3} \leq 15 x_{1} \geq 0, x_{2} \geq 0, x_{3} \geq 0$

We know that for the primal problem

$Max Z_{x} = c x st A x \leq b x \geq 0$

Then the dual problem is

$Min Z_{y} = b^{T} y s . t A^{T} y \leq c^{T} y \geq 0$

Where $T$ denote the transpose

Now comparing with the standard form

$c = (7, 3, 8) A = [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 7 & 1 & 2 \end{matrix}] b = [\begin{matrix} 6 \\ 3 \\ 15 \end{matrix}] ∴ c^{T} = [\begin{matrix} 7 \\ 3 \\ 8 \end{matrix}] A^{T} = [\begin{matrix} 1 & 1 & 7 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{matrix}] b^{T} = (6, 3, 15)$