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To solve the linear programming problem of maximizing [tex]\( z = x + 2y \)[/tex] subject to the given constraints, we can follow these steps:
### Step 1: Write down the constraints
We are given the following constraints:
1. [tex]\( 3x + 4y \leq 36 \)[/tex]
2. [tex]\( 6x + 2y \leq 36 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 2: Identify the feasible region
The feasible region is determined by the intersection of the constraint inequalities in the [tex]\((x,y)\)[/tex]-plane.
### Step 3: Find the vertices of the feasible region
To find the vertices, we solve the system of inequalities where the lines intersect.
First, solve [tex]\( 3x + 4y = 36 \)[/tex]:
1. When [tex]\( x = 0 \)[/tex], [tex]\( 4y = 36 \rightarrow y = 9 \)[/tex].
2. When [tex]\( y = 0 \)[/tex], [tex]\( 3x = 36 \rightarrow x = 12 \)[/tex].
So, the line intersects the axes at points [tex]\( (0, 9) \)[/tex] and [tex]\( (12, 0) \)[/tex].
Now, solve [tex]\( 6x + 2y = 36 \)[/tex]:
1. When [tex]\( x = 0 \)[/tex], [tex]\( 2y = 36 \rightarrow y = 18 \)[/tex].
2. When [tex]\( y = 0 \)[/tex], [tex]\( 6x = 36 \rightarrow x = 6 \)[/tex].
So, the line intersects the axes at points [tex]\( (0, 18) \)[/tex] and [tex]\( (6, 0) \)[/tex].
Next, find the intersection of [tex]\( 3x + 4y = 36 \)[/tex] and [tex]\( 6x + 2y = 36 \)[/tex]:
[tex]\[ \begin{cases} 3x + 4y = 36 \\ 6x + 2y = 36 \end{cases} \][/tex]
Multiply the first equation by 2:
[tex]\( 6x + 8y = 72 \)[/tex]
Subtract the second equation:
[tex]\( 6x + 8y - 6x - 2y = 72 - 36 \)[/tex]
[tex]\( 6y = 36 \rightarrow y = 6 \)[/tex]
Substitute [tex]\( y = 6 \)[/tex] back into [tex]\( 3x + 4y = 36 \)[/tex]:
[tex]\( 3x + 4(6) = 36 \rightarrow 3x + 24 = 36 \rightarrow 3x = 12 \rightarrow x = 4 \)[/tex]
So, another vertex is [tex]\( (4, 6) \)[/tex].
### Step 4: Evaluate [tex]\( z = x + 2y \)[/tex] at each vertex
The vertices of the feasible region are [tex]\( (0, 9) \)[/tex], [tex]\( (12, 0) \)[/tex], [tex]\( (0, 18) \)[/tex], and [tex]\( (4, 6) \)[/tex].
Evaluate [tex]\( z \)[/tex] at these points:
1. At [tex]\( (0, 9) \)[/tex]: [tex]\( z = 0 + 2(9) = 18 \)[/tex]
2. At [tex]\( (12, 0) \)[/tex]: [tex]\( z = 12 + 2(0) = 12 \)[/tex]
3. At [tex]\( (4, 6) \)[/tex]: [tex]\( z = 4 + 2(6) = 4 + 12 = 16 \)[/tex]
4. At [tex]\( (0, 18) \)[/tex]: This point is not in the feasible region because it does not satisfy [tex]\( 3x + 4y \leq 36 \)[/tex].
### Step 5: Determine the maximum value
The maximum value of [tex]\( z \)[/tex] is [tex]\( 18 \)[/tex] at the point [tex]\( (0, 9) \)[/tex].
### Solution
[tex]\[ \text{Maximum is } 18 \text{ at,} \][/tex]
[tex]\[ \begin{array}{l} x = 0 \\ y = 9 \end{array} \][/tex]
### Step 1: Write down the constraints
We are given the following constraints:
1. [tex]\( 3x + 4y \leq 36 \)[/tex]
2. [tex]\( 6x + 2y \leq 36 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 2: Identify the feasible region
The feasible region is determined by the intersection of the constraint inequalities in the [tex]\((x,y)\)[/tex]-plane.
### Step 3: Find the vertices of the feasible region
To find the vertices, we solve the system of inequalities where the lines intersect.
First, solve [tex]\( 3x + 4y = 36 \)[/tex]:
1. When [tex]\( x = 0 \)[/tex], [tex]\( 4y = 36 \rightarrow y = 9 \)[/tex].
2. When [tex]\( y = 0 \)[/tex], [tex]\( 3x = 36 \rightarrow x = 12 \)[/tex].
So, the line intersects the axes at points [tex]\( (0, 9) \)[/tex] and [tex]\( (12, 0) \)[/tex].
Now, solve [tex]\( 6x + 2y = 36 \)[/tex]:
1. When [tex]\( x = 0 \)[/tex], [tex]\( 2y = 36 \rightarrow y = 18 \)[/tex].
2. When [tex]\( y = 0 \)[/tex], [tex]\( 6x = 36 \rightarrow x = 6 \)[/tex].
So, the line intersects the axes at points [tex]\( (0, 18) \)[/tex] and [tex]\( (6, 0) \)[/tex].
Next, find the intersection of [tex]\( 3x + 4y = 36 \)[/tex] and [tex]\( 6x + 2y = 36 \)[/tex]:
[tex]\[ \begin{cases} 3x + 4y = 36 \\ 6x + 2y = 36 \end{cases} \][/tex]
Multiply the first equation by 2:
[tex]\( 6x + 8y = 72 \)[/tex]
Subtract the second equation:
[tex]\( 6x + 8y - 6x - 2y = 72 - 36 \)[/tex]
[tex]\( 6y = 36 \rightarrow y = 6 \)[/tex]
Substitute [tex]\( y = 6 \)[/tex] back into [tex]\( 3x + 4y = 36 \)[/tex]:
[tex]\( 3x + 4(6) = 36 \rightarrow 3x + 24 = 36 \rightarrow 3x = 12 \rightarrow x = 4 \)[/tex]
So, another vertex is [tex]\( (4, 6) \)[/tex].
### Step 4: Evaluate [tex]\( z = x + 2y \)[/tex] at each vertex
The vertices of the feasible region are [tex]\( (0, 9) \)[/tex], [tex]\( (12, 0) \)[/tex], [tex]\( (0, 18) \)[/tex], and [tex]\( (4, 6) \)[/tex].
Evaluate [tex]\( z \)[/tex] at these points:
1. At [tex]\( (0, 9) \)[/tex]: [tex]\( z = 0 + 2(9) = 18 \)[/tex]
2. At [tex]\( (12, 0) \)[/tex]: [tex]\( z = 12 + 2(0) = 12 \)[/tex]
3. At [tex]\( (4, 6) \)[/tex]: [tex]\( z = 4 + 2(6) = 4 + 12 = 16 \)[/tex]
4. At [tex]\( (0, 18) \)[/tex]: This point is not in the feasible region because it does not satisfy [tex]\( 3x + 4y \leq 36 \)[/tex].
### Step 5: Determine the maximum value
The maximum value of [tex]\( z \)[/tex] is [tex]\( 18 \)[/tex] at the point [tex]\( (0, 9) \)[/tex].
### Solution
[tex]\[ \text{Maximum is } 18 \text{ at,} \][/tex]
[tex]\[ \begin{array}{l} x = 0 \\ y = 9 \end{array} \][/tex]
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