MHB Solutions of the given linear programming problem

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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

- - - Updated - - -

I like Serena said:
Looks good! ;)

EDIT: Aargh, overtaken by Opalg.

Nice! Thank you! :o
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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