Consider the two inequalities 3x + 2y ≥ 4
9x + 8y ≤ 14
2y ≥ -3x + 4
2y/2 ≥ -3x/2 + 4/2
y ≥ -3x/2 + 4/2
y ≥ -3x/2 + 2
slope -3/2
y- intercept 2
The black line is y ≥ -3x/2 + 2
8y ≤ -9x + 14
8y/8 ≤ - 9x/8 + 14/8
y ≤ -9x/8 + 14/8
y ≤ - 9x/8 + 7/4
slope -9/8
y- intercept 7/4
The red line is y ≤ - 9x/8 + 7/4
Both points intercect at (2/3, 1)
Find the feasible region for these two inequalities and find its corner. Can anyone tell me what the corner points are?
With two inequalities, you can have a maximum of one intersection point. The feasible region is therefore open.
The feasible region is below the red line, and above the black line.
The corner point has already been found correctly, ... by you!
Ok thanks so I did answer the question correctly.
You're welcome!
To find the feasible region, we need to shade the area that satisfies both inequalities.
First, let's graph the lines represented by the inequalities:
- The black line is y ≥ -3x/2 + 2
- The red line is y ≤ -9x/8 + 7/4
Now, we need to determine which side of each line satisfies the inequality.
For the black line, since it is a "greater than or equal to" inequality, we shade the area above the line.
For the red line, since it is a "less than or equal to" inequality, we shade the area below the line.
Now, we can see the shaded area where both inequalities are satisfied.
To determine the corner points of the feasible region, we look for the vertices of the shaded region.
By visually inspecting the graph, it seems that the corner points are:
1. (2/3, 1)
2. (0, 2)
3. (4/3, 0)