MHB Solutions of the given linear programming problem

AI Thread Summary
The discussion revolves around solving a linear programming problem using the simplex method and the Two-Phase method. The user proposes a solution indicating infinite solutions represented by the equation (x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m) with a minimum value of 8. Another participant confirms the correctness of this solution by demonstrating that the inequalities lead to a minimum condition satisfied by setting z=0. The conversation concludes with mutual acknowledgment of the solution's validity.
mathmari
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Hello! :o
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?
 
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mathmari said:
Hello! :o
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?
Your answer is certainly correct, because if you add the two inequalities you get $2x + 3y + 5z + 4w \geqslant 5+3=8$. Therefore $2x + 3y + 6z + 4w \geqslant 8+z$, which is minimised by taking $z=0$. Your solutions, with $z=0$, clearly satisfy all the given conditions, so they must be right.
 
mathmari said:
Hello! :o
Given the following linear programming problem
$$\min(2x + 3y + 6z + 4w)$$
$$x+2y+3z+w \geq 5$$
$$x+y+2z+3w \geq 3$$
$$x,y,z,w \geq 0$$
I am asked to find all the solutions using the simplex method.

To solve this problem we use the Two-Phase method, don't we?
Then I found that there are infinite many solutions, $(x,y,z,w)=(1-m, 2+0.4m, 0, 0.2m), 0≤m≤1$ with $min=8$.
Could you tell me if this is right?

Looks good! ;)

EDIT: Aargh, overtaken by Opalg.
 
Opalg said:
Your answer is certainly correct, because if you add the two inequalities you get $2x + 3y + 5z + 4w \geqslant 5+3=8$. Therefore $2x + 3y + 6z + 4w \geqslant 8+z$, which is minimised by taking $z=0$. Your solutions, with $z=0$, clearly satisfy all the given conditions, so they must be right.

Great! Thank you for your answer! :o

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I like Serena said:
Looks good! ;)

EDIT: Aargh, overtaken by Opalg.

Nice! Thank you! :o
 
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