Graphing

[johnsons44281]

Hi! I need assistance with a set of 4 equations that I am supposed to graph and find the feasible region indicating all corner points. I continue to second guess myself on a couple of the equations and I guess I am looking for someone to confirm for me if I have resolved these correctly or not because my graph does not seem to be making sense to me. If I can at least confirm my work is correct, then maybe the graph will work itself out. I hope you can help.

1. x >= 0 (>= is greater than or equal to)
This equation graphs in a vertical line directly along the y axis, yes? And my shading would be to the right since the equation is > and I want to shade positive numbers?

2. y >= 2
This equation graphs in a horizontal line intersecting at the second point on the y axis? And because the equation is > I want to shade above this horizontal line?

3. -5x + y <= 14
Using substitution of 0, 1, -2, -3 for x (work below) I get the following points: (0,14), (1,19), (-2,4), (-3,-1)
-5(0) + y <= 14
0 + y <= 14
y <=14

-5(1) + y <=14
-5 + y <= 14
+5 +5
y <=19

-5(-2) + y <= 14
10 + y <= 14
-10 -10
y <= 4

-5(-3) + y <= 14
15 + y <= 14
-15 -15
y <= -1

 

[Deleted member 4993]

Hi! I need assistance with a set of 4 equations that I am supposed to graph and find the feasible region indicating all corner points. I continue to second guess myself on a couple of the equations and I guess I am looking for someone to confirm for me if I have resolved these correctly or not because my graph does not seem to be making sense to me. If I can at least confirm my work is correct, then maybe the graph will work itself out. I hope you can help.

1. x >= 0 (>= is greater than or equal to)
This equation graphs in a vertical line directly along the y axis, yes? And my shading would be to the right since the equation is > and I want to shade positive numbers?

2. y >= 2
This equation graphs in a horizontal line intersecting at the second point on the y axis? And because the equation is > I want to shade above this horizontal line?

3. -5x + y <= 14

Do you know how to graph

y = 5x + 14

If you don't - go to:

http://www.purplemath.com/modules/graphlin.htm

The points of interest would be above this line (for the > sign) and including the line (for the = sign)



Using substitution of 0, 1, -2, -3 for x (work below) I get the following points: (0,14), (1,19), (-2,4), (-3,-1)
-5(0) + y <= 14
0 + y <= 14
y <=14

-5(1) + y <=14
-5 + y <= 14
+5 +5
y <=19

-5(-2) + y <= 14
10 + y <= 14
-10 -10
y <= 4

-5(-3) + y <= 14
15 + y <= 14
-15 -15
y <= -1

 

[mmm4444bot]

Hi! I need assistance with a set of 4 equations that I am supposed to graph and find the feasible region indicating all corner points. I continue to second guess myself on a couple of the equations and I guess I am looking for someone to confirm for me if I have resolved these correctly or not because my graph does not seem to be making sense to me. If I can at least confirm my work is correct, then maybe the graph will work itself out. I hope you can help.

1. x >= 0 (>= is greater than or equal to)
This equation graphs in a vertical line directly along the y axis, yes?

Yes, if you're referring to the equation x = 0.

x ? 0 is not an equation; it is an inequality, so be careful with your terminology.

And my shading would be to the right since the equation is > and I want to shade positive numbers?

You are correct by shading the half of the xy-plane that lies to the right of the y-axis. Realize also that all of the points that make up the y-axis itself are also solutions to the inequality because the inequality sign is greater than OR EQUAL TO. This means that the line you draw on top of the y-axis must be a solid line.

2. y >= 2
This equation graphs in a horizontal line intersecting at the second point on the y axis?

I know what you mean, and you're correct; however, again the terminology is lacking. The equation is y=2. We cannot really refer to the point (0,2) as "the second point on the y-axis". There are an infinite number of points on the y-axis between the origin and ANY y-intercept. I'm keen enough to realize that you're talking about points with whole-numbered coordinates, but try to be more precise in your terminology; others may not be so keen.

And because the equation is > I want to shade above this horizontal line?

Correct, again. Also, again, the line y = 2 is graphed as a solid line because those points are part of the inequality solution.

3. -5x + y <= 14

Using substitution of 0, 1, -2, -3 for x (work below) I get the following points: (0,14), (1,19), (-2,4), (-3,-1)

-5(0) + y <= 14
0 + y <= 14
y <=14

-5(1) + y <=14
-5 + y <= 14
+5 +5
y <=19

-5(-2) + y <= 14
10 + y <= 14
-10 -10
y <= 4

-5(-3) + y <= 14
15 + y <= 14
-15 -15
y <= -1

I'm not sure why you're evaluating these points. You only need two points to graph the line

-5x + y = 14

Since the inequality symbol is less than OR EQUAL TO, you draw a solid line.

Pick any point above or below this line, and put the x- and y-coordinate into the original inequality as a test. If these values make the inequality statement true, then the test point lies in the half of the xy-plane that needs to be shaded. If the test point's coordinates result in a false statement, then you want to shade the other side of the line.

Lastly, I see only 3 inequalities. I see nothing to bound the right side of the region. In other words, you will have three vertices for which you can list the coordinates, but this region will not be a closed shape.

Thanks for showing your work. Let us know if you have more questions.

~ Mark