Sure, let's graph the feasible region for the system of inequalities step by step.
1. Plot the lines for each inequality:
- For the inequality [tex]\( x + y \leq 12 \)[/tex]:
We can start by finding the intercepts.
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 12 \)[/tex] (point: [tex]\( (0, 12) \)[/tex])
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = 12 \)[/tex] (point: [tex]\( (12, 0) \)[/tex])
These points help us draw the line [tex]\( x + y = 12 \)[/tex]. The region below this line is part of the feasible region.
- For the inequality [tex]\( -x + y \leq 6 \)[/tex]:
We can start by finding the intercepts.
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 6 \)[/tex] (point: [tex]\( (0, 6) \)[/tex])
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = -6 \)[/tex] (point: [tex]\( (6, 0) \)[/tex])
However, since [tex]\( x \)[/tex] must be positive, we can ignore the left side beyond the y-axis and focus only on feasible positive values of [tex]\( x \)[/tex].
- For the inequality [tex]\( x \geq 0 \)[/tex]:
This inequality indicates that we are considering the region on the right-hand side of the y-axis.
- For the inequality [tex]\( y \geq 0 \)[/tex]:
This inequality indicates that we are considering the region above the x-axis.
2. Find the corner points of the feasible region:
Since it's a system of linear inequalities, the boundary lines intersect at various points, which are the vertices of the feasible region. We need to find these intersection points.
- The point of intersection between [tex]\( x + y = 12 \)[/tex] and [tex]\( -x + y = 6 \)[/tex] can be found by solving the equations simultaneously:
[tex]\[
x + y = 12 \quad \text{(1)}
\][/tex]
[tex]\[
-x + y = 6 \quad \text{(2)}
\][/tex]
Adding these equations, we get:
[tex]\[
(x + y) + (-x + y) = 12 + 6
\][/tex]
[tex]\[
2y = 18 \implies y = 9
\][/tex]
Substituting [tex]\( y = 9 \)[/tex] back into equation (1):
[tex]\[
x + 9 = 12 \implies x = 3
\][/tex]
Therefore, one intersection point is [tex]\( (3, 9) \)[/tex].
- Other intersection points are at the axes:
[tex]\( (0, 0) \)[/tex], [tex]\( (0, 6) \)[/tex], and [tex]\( (12, 0) \)[/tex]
3. Draw the feasible region:
Now, plot these points: [tex]\( (0, 0) \)[/tex], [tex]\( (0, 6) \)[/tex], [tex]\( (3, 9) \)[/tex], and [tex]\( (12, 0) \)[/tex].
Connect them to form the polygon:
- Start at [tex]\( (0, 0) \)[/tex]
- Move to [tex]\( (0, 6) \)[/tex]
- Then move to [tex]\( (3, 9) \)[/tex]
- Then move to [tex]\( (12, 0) \)[/tex]
- Close the polygon by moving back to [tex]\( (0, 0) \)[/tex]
The resulting shape is a polygon with vertices at [tex]\( (0, 0) \)[/tex], [tex]\( (0, 6) \)[/tex], [tex]\( (3, 9) \)[/tex], and [tex]\( (12, 0) \)[/tex]. This polygon defines the feasible region for the given system of inequalities.
