Consider the two inequalities 3x + 2y ≥ 4

Question

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?

Answer 1

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!

Answer 2

Ok thanks so I did answer the question correctly.

Answer 3

You're welcome!