IDNLearn.com makes it easy to find accurate answers to your questions. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
Sagot :
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].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
