Answer :
To determine the vertices of the feasible region given the constraints:
[tex]\[ \begin{array}{l} x + 3y \leq 6 \\ 4x + 6y \geq 9 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
Let's process each constraint to find the points of intersection:
1. The equation [tex]\(x + 3y = 6\)[/tex]:
- Intersecting [tex]\(x\)[/tex]-axis ([tex]\(y = 0\)[/tex]):
[tex]\[ x = 6 \][/tex]
This gives the point [tex]\((6, 0)\)[/tex].
- Intersecting [tex]\(y\)[/tex]-axis ([tex]\(x = 0\)[/tex]):
[tex]\[ 3y = 6 \implies y = 2 \][/tex]
This gives the point [tex]\((0, 2)\)[/tex].
2. The equation [tex]\(4x + 6y = 9\)[/tex]:
- Intersecting [tex]\(x\)[/tex]-axis ([tex]\(y = 0\)[/tex]):
[tex]\[ 4x = 9 \implies x = \frac{9}{4} = 2.25 \][/tex]
This gives the point [tex]\((2.25, 0)\)[/tex].
- Intersecting [tex]\(y\)[/tex]-axis ([tex]\(x = 0\)[/tex]):
[tex]\[ 6y = 9 \implies y = 1.5 \][/tex]
This gives the point [tex]\((0, 1.5)\)[/tex].
3. Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(4x + 6y = 9\)[/tex]:
- To find the intersection, solve these equations simultaneously.
[tex]\[ \begin{align*} x + 3y &= 6 \quad \text{(equation 1)} \\ 4x + 6y &= 9 \quad \text{(equation 2)} \end{align*} \][/tex]
Multiplying equation 1 by 2 to match [tex]\(y\)[/tex] coefficients:
[tex]\[ \begin{align*} 2x + 6y &= 12 \quad \text{(equation 3)} \\ 4x + 6y &= 9 \quad \text{(equation 2)} \end{align*} \][/tex]
Subtract equation 2 from equation 3:
[tex]\[ -2x = 3 \implies x = -1.5 \][/tex]
Substituting [tex]\(x = -1.5\)[/tex] back into equation 1:
[tex]\[ -1.5 + 3y = 6 \implies 3y = 7.5 \implies y = 2.5 \][/tex]
Hence, the intersection is [tex]\((-1.5, 2.5)\)[/tex].
4. We also have the origin point [tex]\((0, 0)\)[/tex].
Next, we identify the feasible vertices by ensuring they satisfy all constraints.
- Points: [tex]\((6, 0)\)[/tex], [tex]\((0, 2)\)[/tex], [tex]\((2.25, 0)\)[/tex], [tex]\((0, 1.5)\)[/tex], [tex]\((-1.5, 2.5)\)[/tex], and [tex]\((0, 0)\)[/tex].
- Limits: [tex]\(x \geq 0\)[/tex], [tex]\(y \geq 0\)[/tex], [tex]\(x + 3y \leq 6\)[/tex], and [tex]\(4x + 6y \geq 9\)[/tex].
### Checking Feasibility:
1. [tex]\((6, 0)\)[/tex]:
[tex]\[ 6 + 30 = 6 \text{ (True)} \quad 46 + 6*0 \geq 9 \text{ (True)} \][/tex]
2. [tex]\((0, 2)\)[/tex]:
[tex]\[ 0 + 32 = 6 \text{ (True)} \quad 40 + 6*2 = 12 \ (True) \][/tex]
3. [tex]\((2.25, 0)\)[/tex]:
[tex]\[ 2.25 + 30 = 2.25 \ (True) \quad 42.25 + 6*0 = 9 (True) \][/tex]
4. [tex]\((0, 1.5)\)[/tex]:
[tex]\[ 0 + 31.5 = 4.5 (True) \quad 40 + 6*1.5 = 9 (True) \][/tex]
5. [tex]\((0, 0)\)[/tex]:
[tex]\[ 0 + 30 = 0 (True) \quad 40 + 6*0 = 0(False) \][/tex]
6. [tex]\((-1.5, 2.5)\)[/tex]:
[tex]\[ -1.5 + 3*2.5= (4.5+ (-1.5)) <= 6 (False) \quad 4x + 6y= 9 (True) \][/tex]
We reject this infeasible point.
Valid feasible vertices for region:
[tex]\[ (0, 2), (0, 1.5), (6, 0), (2.25, 0) \][/tex]
Thus, the vertices of the feasible region are [tex]\((0, 2)\)[/tex], [tex]\((0, 1.5)\)[/tex], [tex]\((6, 0)\)[/tex], and [tex]\((2.25, 0)\)[/tex].
[tex]\[ \begin{array}{l} x + 3y \leq 6 \\ 4x + 6y \geq 9 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
Let's process each constraint to find the points of intersection:
1. The equation [tex]\(x + 3y = 6\)[/tex]:
- Intersecting [tex]\(x\)[/tex]-axis ([tex]\(y = 0\)[/tex]):
[tex]\[ x = 6 \][/tex]
This gives the point [tex]\((6, 0)\)[/tex].
- Intersecting [tex]\(y\)[/tex]-axis ([tex]\(x = 0\)[/tex]):
[tex]\[ 3y = 6 \implies y = 2 \][/tex]
This gives the point [tex]\((0, 2)\)[/tex].
2. The equation [tex]\(4x + 6y = 9\)[/tex]:
- Intersecting [tex]\(x\)[/tex]-axis ([tex]\(y = 0\)[/tex]):
[tex]\[ 4x = 9 \implies x = \frac{9}{4} = 2.25 \][/tex]
This gives the point [tex]\((2.25, 0)\)[/tex].
- Intersecting [tex]\(y\)[/tex]-axis ([tex]\(x = 0\)[/tex]):
[tex]\[ 6y = 9 \implies y = 1.5 \][/tex]
This gives the point [tex]\((0, 1.5)\)[/tex].
3. Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(4x + 6y = 9\)[/tex]:
- To find the intersection, solve these equations simultaneously.
[tex]\[ \begin{align*} x + 3y &= 6 \quad \text{(equation 1)} \\ 4x + 6y &= 9 \quad \text{(equation 2)} \end{align*} \][/tex]
Multiplying equation 1 by 2 to match [tex]\(y\)[/tex] coefficients:
[tex]\[ \begin{align*} 2x + 6y &= 12 \quad \text{(equation 3)} \\ 4x + 6y &= 9 \quad \text{(equation 2)} \end{align*} \][/tex]
Subtract equation 2 from equation 3:
[tex]\[ -2x = 3 \implies x = -1.5 \][/tex]
Substituting [tex]\(x = -1.5\)[/tex] back into equation 1:
[tex]\[ -1.5 + 3y = 6 \implies 3y = 7.5 \implies y = 2.5 \][/tex]
Hence, the intersection is [tex]\((-1.5, 2.5)\)[/tex].
4. We also have the origin point [tex]\((0, 0)\)[/tex].
Next, we identify the feasible vertices by ensuring they satisfy all constraints.
- Points: [tex]\((6, 0)\)[/tex], [tex]\((0, 2)\)[/tex], [tex]\((2.25, 0)\)[/tex], [tex]\((0, 1.5)\)[/tex], [tex]\((-1.5, 2.5)\)[/tex], and [tex]\((0, 0)\)[/tex].
- Limits: [tex]\(x \geq 0\)[/tex], [tex]\(y \geq 0\)[/tex], [tex]\(x + 3y \leq 6\)[/tex], and [tex]\(4x + 6y \geq 9\)[/tex].
### Checking Feasibility:
1. [tex]\((6, 0)\)[/tex]:
[tex]\[ 6 + 30 = 6 \text{ (True)} \quad 46 + 6*0 \geq 9 \text{ (True)} \][/tex]
2. [tex]\((0, 2)\)[/tex]:
[tex]\[ 0 + 32 = 6 \text{ (True)} \quad 40 + 6*2 = 12 \ (True) \][/tex]
3. [tex]\((2.25, 0)\)[/tex]:
[tex]\[ 2.25 + 30 = 2.25 \ (True) \quad 42.25 + 6*0 = 9 (True) \][/tex]
4. [tex]\((0, 1.5)\)[/tex]:
[tex]\[ 0 + 31.5 = 4.5 (True) \quad 40 + 6*1.5 = 9 (True) \][/tex]
5. [tex]\((0, 0)\)[/tex]:
[tex]\[ 0 + 30 = 0 (True) \quad 40 + 6*0 = 0(False) \][/tex]
6. [tex]\((-1.5, 2.5)\)[/tex]:
[tex]\[ -1.5 + 3*2.5= (4.5+ (-1.5)) <= 6 (False) \quad 4x + 6y= 9 (True) \][/tex]
We reject this infeasible point.
Valid feasible vertices for region:
[tex]\[ (0, 2), (0, 1.5), (6, 0), (2.25, 0) \][/tex]
Thus, the vertices of the feasible region are [tex]\((0, 2)\)[/tex], [tex]\((0, 1.5)\)[/tex], [tex]\((6, 0)\)[/tex], and [tex]\((2.25, 0)\)[/tex].
