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A furniture company displays bedroom sets which require 21 square meters of space and living room sets which require 42 square...Asked by Lucy
A furniture company displays bedroom sets which require 21 square meters of space and living room sets which require 42 square meters of space. The company,which has 546 square meters of available space, wants to display at least 6 bedroom sets and at least 5 living room sets.
1) Let x represent the number of bedroom suits and y represent the number of living room suits. Write a system of inequalities to represent the number of furniture sets that can be displayed. I think it is 21x + 42y <= 546. Is this correct.
2) Draw a graph showing the feasible region. Label the coordinates of the vertices of the feasible region. I am not sure how to do this--the example in the book does not explain how to do this.
1) Let x represent the number of bedroom suits and y represent the number of living room suits. Write a system of inequalities to represent the number of furniture sets that can be displayed. I think it is 21x + 42y <= 546. Is this correct.
2) Draw a graph showing the feasible region. Label the coordinates of the vertices of the feasible region. I am not sure how to do this--the example in the book does not explain how to do this.
Answers
Answered by
Reiny
You are on the right track,
graph 21x + 42y ≤ 546 , (the x intercept is 26 and the y intercept is 13)
Don't forget the other two restrictions,
x ≤ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 <b>AND</b> x ≤ 6 <b>AND</b> y ≥ 5
I see a right-angled triangle
BTW, since x and y have to be whole numbers, the intersection points would not be feasible solution points, you will have to go the next set of integer values that satisfy the above conditions.
BTW#2, somebody posted this same question a few weeks ago, but they also included additonal data for the cost of the bedrooms and living rooms
graph 21x + 42y ≤ 546 , (the x intercept is 26 and the y intercept is 13)
Don't forget the other two restrictions,
x ≤ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 <b>AND</b> x ≤ 6 <b>AND</b> y ≥ 5
I see a right-angled triangle
BTW, since x and y have to be whole numbers, the intersection points would not be feasible solution points, you will have to go the next set of integer values that satisfy the above conditions.
BTW#2, somebody posted this same question a few weeks ago, but they also included additonal data for the cost of the bedrooms and living rooms
Answered by
Reiny - Correction
"Don't forget the other two restrictions,
x ≤ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 AND x ≤ 6 AND y ≥ 5"
should have been
Don't forget the other two restrictions,
x ≥ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 AND x ≥ 6 AND y ≥ 5
x ≤ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 AND x ≤ 6 AND y ≥ 5"
should have been
Don't forget the other two restrictions,
x ≥ 6 and y ≥ 5
so you are looking at the region
21x + 42y ≤ 546 AND x ≥ 6 AND y ≥ 5
Answered by
bobpursley
Make the graph as follows:
Horizontal axis: bedroom sets
Vertical axis: living room sets
Now, if all the area was living room sets, it would have the coordinate (546/42,0) so plot that point. Then plot the point of all bedroom sets (0,546/21)
Connect the points with a line. Now draw the constrains
Vertical line at x (bedroom)=6
Vertical line at x=0
horizontal line at y=0
horizontal line at y=5
The feasible area is enclosed by the lines.
Horizontal axis: bedroom sets
Vertical axis: living room sets
Now, if all the area was living room sets, it would have the coordinate (546/42,0) so plot that point. Then plot the point of all bedroom sets (0,546/21)
Connect the points with a line. Now draw the constrains
Vertical line at x (bedroom)=6
Vertical line at x=0
horizontal line at y=0
horizontal line at y=5
The feasible area is enclosed by the lines.
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