Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To solve this problem, let’s follow these steps:
1. Graph the constraints: We will graph the feasible region determined by the constraints.
2. Identify the corner points: Determine the intersection points of the constraints.
3. Evaluate the objective function at the corner points: Calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point.
4. Determine the maximum value: Identify which point gives the maximum value for the objective function [tex]\(C\)[/tex].
### Step 1: Graph the Constraints
The given constraints are:
1. [tex]\(x \geq 0\)[/tex]
2. [tex]\(y \geq 0\)[/tex]
3. [tex]\(y \leq \frac{1}{5}x + 2\)[/tex]
4. [tex]\(y + x \leq 5\)[/tex]
### Step 2: Identify the Intersection Points
To find the intersection points, we solve the equations derived from the constraints:
#### Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex]:
Set [tex]\(y = \frac{1}{5}x + 2\)[/tex] into [tex]\( y + x = 5\)[/tex]:
[tex]\[ \frac{1}{5}x + 2 + x = 5 \][/tex]
[tex]\[ x + \frac{1}{5}x = 3 \][/tex]
[tex]\[ \frac{6}{5}x = 3 \][/tex]
[tex]\[ x = \frac{5 \cdot 3}{6} = 2.5 \][/tex]
[tex]\[ y = \frac{1}{5}(2.5) + 2 = 2.5 \][/tex]
So, one of the intersection points is [tex]\((2.5, 2.5)\)[/tex].
#### Other points from constraints:
- [tex]\( (0, 2) \)[/tex] : Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\(x = 0 \)[/tex]
- [tex]\( (5, 0) \)[/tex] : Intersection of [tex]\( y + x = 5 \)[/tex] and [tex]\(y = 0 \)[/tex]
- [tex]\( (0, 0) \)[/tex] : Intersection of [tex]\( y = 0 \)[/tex] and [tex]\(x = 0 \)[/tex]
So, the corner points of the feasible region are:
[tex]\[ (2.5, 2.5), (0, 2), (0, 0), (5, 0) \][/tex]
### Step 3: Evaluate the Objective Function at Each Corner Point
We now calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point:
- At [tex]\((2.5, 2.5)\)[/tex]:
[tex]\[ C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10 \][/tex]
- At [tex]\((0, 2)\)[/tex]:
[tex]\[ C = 7(0) - 3(2) = 0 - 6 = -6 \][/tex]
- At [tex]\((0, 0)\)[/tex]:
[tex]\[ C = 7(0) - 3(0) = 0 \][/tex]
- At [tex]\((5, 0)\)[/tex]:
[tex]\[ C = 7(5) - 3(0) = 35 - 0 = 35 \][/tex]
### Step 4: Determine the Maximum Value
Comparing the values obtained:
[tex]\[ 10, -6, 0, 35 \][/tex]
The maximum value of the objective function is 35 which occurs at the point [tex]\((5, 0)\)[/tex].
### Conclusion
Therefore, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that maximize the objective function [tex]\(C = 7x - 3y\)[/tex] within the given constraints are:
[tex]\[ x = 5 \quad \text{and} \quad y = 0 \][/tex]
Thus, the maximum value of the objective function is 35 at the point [tex]\((5, 0)\)[/tex].
1. Graph the constraints: We will graph the feasible region determined by the constraints.
2. Identify the corner points: Determine the intersection points of the constraints.
3. Evaluate the objective function at the corner points: Calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point.
4. Determine the maximum value: Identify which point gives the maximum value for the objective function [tex]\(C\)[/tex].
### Step 1: Graph the Constraints
The given constraints are:
1. [tex]\(x \geq 0\)[/tex]
2. [tex]\(y \geq 0\)[/tex]
3. [tex]\(y \leq \frac{1}{5}x + 2\)[/tex]
4. [tex]\(y + x \leq 5\)[/tex]
### Step 2: Identify the Intersection Points
To find the intersection points, we solve the equations derived from the constraints:
#### Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex]:
Set [tex]\(y = \frac{1}{5}x + 2\)[/tex] into [tex]\( y + x = 5\)[/tex]:
[tex]\[ \frac{1}{5}x + 2 + x = 5 \][/tex]
[tex]\[ x + \frac{1}{5}x = 3 \][/tex]
[tex]\[ \frac{6}{5}x = 3 \][/tex]
[tex]\[ x = \frac{5 \cdot 3}{6} = 2.5 \][/tex]
[tex]\[ y = \frac{1}{5}(2.5) + 2 = 2.5 \][/tex]
So, one of the intersection points is [tex]\((2.5, 2.5)\)[/tex].
#### Other points from constraints:
- [tex]\( (0, 2) \)[/tex] : Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\(x = 0 \)[/tex]
- [tex]\( (5, 0) \)[/tex] : Intersection of [tex]\( y + x = 5 \)[/tex] and [tex]\(y = 0 \)[/tex]
- [tex]\( (0, 0) \)[/tex] : Intersection of [tex]\( y = 0 \)[/tex] and [tex]\(x = 0 \)[/tex]
So, the corner points of the feasible region are:
[tex]\[ (2.5, 2.5), (0, 2), (0, 0), (5, 0) \][/tex]
### Step 3: Evaluate the Objective Function at Each Corner Point
We now calculate the value of the objective function [tex]\(C = 7x - 3y\)[/tex] at each corner point:
- At [tex]\((2.5, 2.5)\)[/tex]:
[tex]\[ C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10 \][/tex]
- At [tex]\((0, 2)\)[/tex]:
[tex]\[ C = 7(0) - 3(2) = 0 - 6 = -6 \][/tex]
- At [tex]\((0, 0)\)[/tex]:
[tex]\[ C = 7(0) - 3(0) = 0 \][/tex]
- At [tex]\((5, 0)\)[/tex]:
[tex]\[ C = 7(5) - 3(0) = 35 - 0 = 35 \][/tex]
### Step 4: Determine the Maximum Value
Comparing the values obtained:
[tex]\[ 10, -6, 0, 35 \][/tex]
The maximum value of the objective function is 35 which occurs at the point [tex]\((5, 0)\)[/tex].
### Conclusion
Therefore, the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that maximize the objective function [tex]\(C = 7x - 3y\)[/tex] within the given constraints are:
[tex]\[ x = 5 \quad \text{and} \quad y = 0 \][/tex]
Thus, the maximum value of the objective function is 35 at the point [tex]\((5, 0)\)[/tex].
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.