Linear Programming

filterkss

filterkss

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Anyone know linear programming and how to solve this using graphical method? Also is it possible to find the feasible region?

Minimize: 5x + 10y

Subject to: 2x + y ≥ 10
× + y ≥ 15
x ≥ 0 ; y ≥ 0
 
filterkss said:
Anyone know linear programming and how to solve this using graphical method? Also is it possible to find the feasible region?

Minimize: 5x + 10y

Subject to: 2x + y ≥ 10
× + y ≥ 15
x ≥ 0 ; y ≥ 0
Click to expand...
If I were to solve this problem graphically, I would get some graph paper(s), pencil, eraser and a straight edge. Then, on a single graph paper I would plot:

y = -2x + 10
y = -x + 15

These and the x & y axes will define the constraints.

continue......

If you have further questions, please post your work including the graph.
 
Last edited by a moderator:
To add to what the Great Khan said, after graphing the straight lines; you would then test one point from both sides of the graphed lined to see if that side of the line satisfies the inequality from which the line is derived. Shade the side of the line which does not satisfy the inequality. Do this for all the inequalities and you will be left with a region whose enclosed points all satisfy the inequalities.

But since you are looking to find such a point that not only satisfies the inequalities but also minimises the objective function given, you would look at all the vertexes of the unshaded region,( which are actually just the intersections of lines graphed from the inequalities) and substitute their [MATH](x,y)[/MATH] co-ordinates into the objective function you are looking to minimise and record the output they yield.
Then just pick the [MATH](x,y)[/MATH] which yields the lowest value when input into the objective function.
 
Last edited:
Oiler said:
To add to what the Great Khan said, after graphing the straight lines; you would then test one co-ordinate from both sides of the graphed lined to see if that side of the line satisfies the inequality from which the line is derived. Shade the side of the line which does not satisfy the inequality. Do this for all the inequalities and you will be left with a region whose enclosed co-ordinates all satisfy the inequalities.

But since you are looking to find such a co-ordinate that not only satisfies the inequalities but also minimises the objective function given, you would look at all the vertexes of the unshaded region,( which are actually just the intersections of lines graphed from the inequalities) and substitute their [MATH](x,y)[/MATH] co-ordinates into the objective function you are looking to minimise and record the output they yield.
Then just pick the [MATH](x,y)[/MATH] which yields the lowest value when input into the objective function.
Click to expand...
A point has two co-ordinates! You have the x-coordinate and y-coordinate. You should say test a point, not a co-ordinate. Is this clear?
 
Subhotosh Khan said:
If I were to solve this problem graphically, I would get some graph paper(s), pencil, eraser and a straight edge. Then, on a single graph paper I would plot:

y = -2x + 10
y = -x + 15

These and the x & y axes will define the constraints.

continue......

If you have further questions, please post your work including the graph.
Click to expand...

I have al ready plot the x and y and im having difficulty determining shading the feasible region since both the constraints are FALSE 10CB3846-AF14-4493-81C4-33ED250A8215.jpeg
 
I have already dete
filterkss said:
Anyone know linear programming and how to solve this using graphical method? Also is it possible to find the feasible region?

Minimize: 5x + 10y

Subject to: 2x + y ≥ 10
× + y ≥ 15
x ≥ 0 ; y ≥ 0
Click to expand...

I have already determined the x and y now im having difficulty determining the shade regions since the constraints are FALSE
 

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Minimize: 5x + 10y

Subject to: 2x + y ≥ 10
× + y ≥ 15
x ≥ 0 ; y ≥ 0

I’m having difficulty determining the feasible region since the constraints are FALSE

A64F6038-CF10-4D25-B02B-AC625759DB16.jpeg
 
Based on your work, maybe one of the inequalities was x + 3y > 15 and not x+y> 15??

How can the constraints be wrong? They were given to you!
 
Jomo said:
Based on your work, maybe one of the inequalities was x + 3y > 15 and not x+y> 15??

How can the constraints be wrong? They were given to you!
Click to expand...
Oops it was a typo it's
Jomo said:
Based on your work, maybe one of the inequalities was x + 3y > 15 and not x+y> 15??

How can the constraints be wrong? They were given to you!
Click to expand...
Oopss it's a typo yeah it's x + 3y ≥ 15

And no i'm not pertaining that the given constraints are wrong.

What I mean is when i chose point (0,0) and substitute it into the constraints it does not satify the inequality thus it's FALSE and there's nothing to shade.

So I'm asking if there is a possible feasible region here.
 
Yes there is a feasible region.
You need to fix the graph as 1 as one of the lines is not correct. Initially I said both were wrong but then realized that you just had a typo and in fact the line was drawn correctly.

Given just one constraint either all the points on the same side as your test point are included in the inequality OR all the points on the other side of the line are all included in the inequality. So if (0,0) does not satisfy the 1st inequality then shade the other side of the line. Do the same with the 2nd inequality. Look for over lapping and you found your feasible space.
 
filterkss said:
I have already dete


I have already determined the x and y now im having difficulty determining the shade regions since the constraints are FALSE
Click to expand...
1602254741552.png

y = -2x + 10
y = -x + 15

At which point do the two line intersect?

Your sketch is incorrect.

Please fix it and repost (including x and y intercepts). Then we can discuss further.
 
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