# 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.

## 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 AND x ≤ 6 AND 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

## "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

## 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.

## 1) To write a system of inequalities to represent the number of furniture sets that can be displayed, we can start by considering the constraints given. We know that the company wants to display at least 6 bedroom sets and at least 5 living room sets.

Let's use x to represent the number of bedroom sets and y to represent the number of living room sets.

The constraint for the bedroom sets is that they require 21 square meters of space. So, the total space required for the bedroom sets would be 21x.

Similarly, the constraint for the living room sets is that they require 42 square meters of space. So, the total space required for the living room sets would be 42y.

The company has a total available space of 546 square meters, so the total space used by the display should be less than or equal to 546.

Putting all this together, we can write the system of inequalities:

21x + 42y <= 546 (to represent the total space used by the display)
x >= 6 (to represent the minimum number of bedroom sets)
y >= 5 (to represent the minimum number of living room sets)

So, the correct system of inequalities would be 21x + 42y <= 546, x >= 6, and y >= 5.

2) To draw a graph showing the feasible region, we can start by converting the inequalities into equations and graphing them.

First, convert the inequality 21x + 42y <= 546 into an equation: 21x + 42y = 546.

Now, let's solve this equation for both x and y:

1) For x = 0, solving the equation gives us: 21(0) + 42y = 546. From this, we find that y = 13.

2) For y = 0, solving the equation gives us: 21x + 42(0) = 546. From this, we find that x = 26.

Plot these two points (0, 13) and (26, 0) on a graph.

Additionally, plot the points (6, 0) and (0, 5) to represent the minimum number of bedroom and living room sets respectively.

Now, draw a line connecting these points. This line represents the inequality 21x + 42y <= 546.

Shade the region below this line, as this represents the feasible region that satisfies the inequality.

Label the coordinates of the vertices of the feasible region, which are the points where the lines intersect.

The vertices of the feasible region can be found by solving the system of equations formed by the inequalities:

21x + 42y = 546, x = 6, and y = 5.

The feasible region can be labeled accordingly.