To find the vertices of the feasible region given the constraints:
1. [tex]\( x + 3y \leq 6 \)[/tex]
2. [tex]\( 4x + 6y \geq 9 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
We need to determine the points where the boundary lines intersect each other and the coordinate axes [tex]\(x = 0\)[/tex] and [tex]\(y = 0\)[/tex].
### Step-by-Step Solution:
#### Finding Intersections:
1. Intersection of [tex]\( x + 3y = 6 \)[/tex] and [tex]\( y = 0 \)[/tex]:
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( x + 3y = 6 \)[/tex]:
[tex]\[
x + 3(0) = 6 \implies x = 6
\][/tex]
- Point: [tex]\( (6, 0) \)[/tex]
2. Intersection of [tex]\( x = 0 \)[/tex] and [tex]\( x + 3y = 6 \)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( x + 3y = 6 \)[/tex]:
[tex]\[
0 + 3y = 6 \implies y = 2
\][/tex]
- Point: [tex]\( (0, 2) \)[/tex]
3. Intersection of [tex]\( x + 3y = 6 \)[/tex] and [tex]\( 4x + 6y = 9 \)[/tex] (Setting up the system of equations):
[tex]\[
\begin{cases}
x + 3y = 6 \\
4x + 6y = 9
\end{cases}
\][/tex]
- Solving the system leads to the intersection point [tex]\( \left(-\frac{3}{2}, \frac{5}{2}\right) \)[/tex]
#### Checking Feasibility:
4. Intersection with Coordinate Axes:
- Origin [tex]\( (0, 0) \)[/tex]
- By definition [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex], so:
1. [tex]\( (0, 0) \)[/tex]
2. [tex]\( (6, 0) \)[/tex]
3. [tex]\( (0, 2) \)[/tex]
4. [tex]\( \left(-\frac{3}{2}, \frac{5}{2}\right) \)[/tex] - Not within the feasible region as both x and y should be non-negative.
#### Solution:
Combining all feasible points, we get:
[tex]\[
\text{Vertices are: } (0, 0), (6, 0), (0, 2), \left(-\frac{3}{2}, \frac{5}{2}\right)
\][/tex]
But since [tex]\(\left(-\frac{3}{2}, \frac{5}{2}\right)\)[/tex] is not valid in the given constraints,
So we reconsider:
[tex]\[
\text{Vertices are: } (0, 0), (6, 0), (0, 2)
\][/tex]
So, the correct answer is:
[tex]\[
(0,0), (0,2), (6, 0)
\][/tex]
