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Graph the feasible region for the following system of inequalities by drawing a polygon around the feasible region. Click to set the corner points.

[tex]\[ \begin{cases}
x + y \leq 12 \\
-x + y \leq 6 \\
x \geq 0 \\
y \geq 0
\end{cases} \][/tex]

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Draw: [tex]$\square$[/tex]

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Sagot :

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.