Get detailed and reliable answers to your questions on IDNLearn.com. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
To find the maximum profit for the profit function [tex]\( P = 2x + 5y \)[/tex] subject to the constraints:
[tex]\[ \begin{cases} 2x + y \leq 80 \\ x + 2y \leq 80 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we'll follow these steps:
### Step 1: Understand the constraints
We need to visualize the constraints on a graph and find their intersections to determine the feasible region.
- The first constraint [tex]\(2x + y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(y \leq 80\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 40\)[/tex].
- Intersection points can be found by setting one variable to 0 and solving for the other.
- The second constraint [tex]\(x + 2y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(2y \leq 80 \Rightarrow y \leq 40\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 80\)[/tex].
- The non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex] denote that the solutions must be in the first quadrant.
### Step 2: Identify intersection points
To find the vertices of the feasible region, we solve the system of linear equations given by the constraints:
1. Intersection of [tex]\(2x + y = 80\)[/tex] and [tex]\(x + 2y = 80\)[/tex]:
Solving this system of equations:
[tex]\[ \begin{cases} 2x + y = 80 \\ x + 2y = 80 \end{cases} \][/tex]
Multiply the second equation by 2 to eliminate one of the variables:
[tex]\[ \begin{cases} 2x + y = 80 \\ 2x + 4y = 160 \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ 2x + 4y - (2x + y) = 160 - 80 \\ 3y = 80 \\ y = \frac{80}{3} \][/tex]
Substitute [tex]\( y = \frac{80}{3} \)[/tex] back into the first equation [tex]\( 2x + \frac{80}{3} = 80 \)[/tex]:
[tex]\[ 2x = 80 - \frac{80}{3} \\ 2x = \frac{240}{3} - \frac{80}{3} \\ 2x = \frac{160}{3} \\ x = \frac{80}{3} \][/tex]
So, one vertex is [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex].
2. Intersection of [tex]\(2x + y = 80\)[/tex] with the x-axis (where [tex]\( y = 0 \)[/tex]):
Set [tex]\( y = 0 \)[/tex] in the first constraint:
[tex]\[ 2x = 80 \\ x = 40 \][/tex]
So, another vertex is [tex]\( (40, 0) \)[/tex].
3. Intersection of [tex]\(x + 2y = 80\)[/tex] with the y-axis (where [tex]\( x = 0 \)[/tex]):
Set [tex]\( x = 0 \)[/tex] in the second constraint:
[tex]\[ 2y = 80 \\ y = 40 \][/tex]
So, another vertex is [tex]\( (0, 40) \)[/tex].
### Step 3: Calculate the profit at each vertex
Evaluate the profit function [tex]\( P = 2x + 5y \)[/tex] at each vertex of the feasible region:
1. At [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex]:
[tex]\[ P = 2 \cdot \frac{80}{3} + 5 \cdot \frac{80}{3} = \frac{160}{3} + \frac{400}{3} = \frac{560}{3} \approx 186.67 \][/tex]
2. At [tex]\( (40, 0) \)[/tex]:
[tex]\[ P = 2 \cdot 40 + 5 \cdot 0 = 80 \][/tex]
3. At [tex]\( (0, 40) \)[/tex]:
[tex]\[ P = 2 \cdot 0 + 5 \cdot 40 = 200 \][/tex]
### Step 4: Determine the maximum profit
By evaluating the profit function at all vertices, we observe that the maximum profit occurs at [tex]\( (0, 40) \)[/tex] and the maximum profit is 200.
Therefore, the optimal solution is [tex]\( x = 0 \)[/tex] and [tex]\( y = 40 \)[/tex], yielding a maximum profit of 200.
[tex]\[ \begin{cases} 2x + y \leq 80 \\ x + 2y \leq 80 \\ x \geq 0 \\ y \geq 0 \end{cases} \][/tex]
we'll follow these steps:
### Step 1: Understand the constraints
We need to visualize the constraints on a graph and find their intersections to determine the feasible region.
- The first constraint [tex]\(2x + y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(y \leq 80\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 40\)[/tex].
- Intersection points can be found by setting one variable to 0 and solving for the other.
- The second constraint [tex]\(x + 2y \leq 80\)[/tex]:
- If [tex]\(x = 0\)[/tex], then [tex]\(2y \leq 80 \Rightarrow y \leq 40\)[/tex].
- If [tex]\(y = 0\)[/tex], then [tex]\(x \leq 80\)[/tex].
- The non-negativity constraints [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex] denote that the solutions must be in the first quadrant.
### Step 2: Identify intersection points
To find the vertices of the feasible region, we solve the system of linear equations given by the constraints:
1. Intersection of [tex]\(2x + y = 80\)[/tex] and [tex]\(x + 2y = 80\)[/tex]:
Solving this system of equations:
[tex]\[ \begin{cases} 2x + y = 80 \\ x + 2y = 80 \end{cases} \][/tex]
Multiply the second equation by 2 to eliminate one of the variables:
[tex]\[ \begin{cases} 2x + y = 80 \\ 2x + 4y = 160 \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ 2x + 4y - (2x + y) = 160 - 80 \\ 3y = 80 \\ y = \frac{80}{3} \][/tex]
Substitute [tex]\( y = \frac{80}{3} \)[/tex] back into the first equation [tex]\( 2x + \frac{80}{3} = 80 \)[/tex]:
[tex]\[ 2x = 80 - \frac{80}{3} \\ 2x = \frac{240}{3} - \frac{80}{3} \\ 2x = \frac{160}{3} \\ x = \frac{80}{3} \][/tex]
So, one vertex is [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex].
2. Intersection of [tex]\(2x + y = 80\)[/tex] with the x-axis (where [tex]\( y = 0 \)[/tex]):
Set [tex]\( y = 0 \)[/tex] in the first constraint:
[tex]\[ 2x = 80 \\ x = 40 \][/tex]
So, another vertex is [tex]\( (40, 0) \)[/tex].
3. Intersection of [tex]\(x + 2y = 80\)[/tex] with the y-axis (where [tex]\( x = 0 \)[/tex]):
Set [tex]\( x = 0 \)[/tex] in the second constraint:
[tex]\[ 2y = 80 \\ y = 40 \][/tex]
So, another vertex is [tex]\( (0, 40) \)[/tex].
### Step 3: Calculate the profit at each vertex
Evaluate the profit function [tex]\( P = 2x + 5y \)[/tex] at each vertex of the feasible region:
1. At [tex]\( \left(\frac{80}{3}, \frac{80}{3}\right) \)[/tex]:
[tex]\[ P = 2 \cdot \frac{80}{3} + 5 \cdot \frac{80}{3} = \frac{160}{3} + \frac{400}{3} = \frac{560}{3} \approx 186.67 \][/tex]
2. At [tex]\( (40, 0) \)[/tex]:
[tex]\[ P = 2 \cdot 40 + 5 \cdot 0 = 80 \][/tex]
3. At [tex]\( (0, 40) \)[/tex]:
[tex]\[ P = 2 \cdot 0 + 5 \cdot 40 = 200 \][/tex]
### Step 4: Determine the maximum profit
By evaluating the profit function at all vertices, we observe that the maximum profit occurs at [tex]\( (0, 40) \)[/tex] and the maximum profit is 200.
Therefore, the optimal solution is [tex]\( x = 0 \)[/tex] and [tex]\( y = 40 \)[/tex], yielding a maximum profit of 200.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.
