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Sagot :
To solve the given linear programming problem by graphing, we need to follow these steps:
Step 1: Graph the Constraints
---------------------------------
The constraints given in the problem are:
1. [tex]\( x + 2y \leq 10 \)[/tex]
2. [tex]\( 4x + y \leq 12 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
Let's find the intersections of these lines while ensuring non-negative [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
1) [tex]\(x + 2y \leq 10\)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( 2y \leq 10 \implies y \leq 5 \)[/tex]. So, we have the point [tex]\( (0, 5) \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x \leq 10 \)[/tex]. So, we have the point [tex]\( (10, 0) \)[/tex].
2) [tex]\(4x + y \leq 12\)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y \leq 12 \)[/tex]. So, we have the point [tex]\( (0, 12) \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( 4x \leq 12 \implies x \leq 3 \)[/tex]. So, we have the point [tex]\( (3, 0) \)[/tex].
Now, let's find the point of intersection of the two lines:
- [tex]\(x + 2y = 10\)[/tex]
- [tex]\(4x + y = 12\)[/tex]
Solve these equations simultaneously:
Multiply the first equation by 4:
[tex]\(4x + 8y = 40\)[/tex]
Now subtract the second equation from this result:
[tex]\((4x + 8y) - (4x + y) = 40 - 12\)[/tex]
[tex]\(7y = 28\)[/tex]
[tex]\( y = 4 \)[/tex]
Substitute [tex]\(y = 4\)[/tex] back into the first equation:
[tex]\(x + 2*4 = 10\)[/tex]
[tex]\(x + 8 = 10\)[/tex]
[tex]\(x = 2\)[/tex]
So, the point of intersection is [tex]\( (2, 4) \)[/tex].
Step 2: Identify the Feasible Region
--------------------------------------
Plotting the lines on a graph and identifying the region that satisfies all constraints is crucial. The feasible region will be bounded by the lines and the axes.
Vertices of the feasible region:
- [tex]\( (0,5) \)[/tex]
- [tex]\( (10, 0) \)[/tex]
- [tex]\( (3, 0) \)[/tex]
- [tex]\( (0, 12) \)[/tex] (but it is beyond the bound defined by [tex]\(4x+y \leq 12\)[/tex])
- [tex]\( (2, 4) \)[/tex] (intersection of [tex]\( x + 2y = 10 \)[/tex] and [tex]\( 4x + y = 12 \)[/tex])
Step 3: Evaluate the Objective Function at Each Vertex
----------------------------------------------------------
Objective function: [tex]\(P = 7x + 9y\)[/tex]
Calculate [tex]\(P\)[/tex] at each vertex:
- At [tex]\( (0, 5) \)[/tex]:
[tex]\( P = 7(0) + 9(5) = 45 \)[/tex]
- At [tex]\( (10, 0) \)[/tex]:
[tex]\( P = 7(10) + 9(0) = 70 \)[/tex]
- At [tex]\( (3, 0) \)[/tex]:
[tex]\( P = 7(3) + 9(0) = 21 \)[/tex]
- At [tex]\( (2, 4) \)[/tex]:
[tex]\( P = 7(2) + 9(4) = 14 + 36 = 50 \)[/tex]
Step 4: Determine the Maximum Value
-------------------------------------
By evaluating the objective function at each vertex, we observe that the maximum value is [tex]\(P = 50\)[/tex] at point [tex]\( (2, 4) \)[/tex].
Thus,
- [tex]\( x = 2 \)[/tex]
- [tex]\( y = 4 \)[/tex]
- Maximum value [tex]\( P = 50 \)[/tex]
Step 1: Graph the Constraints
---------------------------------
The constraints given in the problem are:
1. [tex]\( x + 2y \leq 10 \)[/tex]
2. [tex]\( 4x + y \leq 12 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
Let's find the intersections of these lines while ensuring non-negative [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
1) [tex]\(x + 2y \leq 10\)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( 2y \leq 10 \implies y \leq 5 \)[/tex]. So, we have the point [tex]\( (0, 5) \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( x \leq 10 \)[/tex]. So, we have the point [tex]\( (10, 0) \)[/tex].
2) [tex]\(4x + y \leq 12\)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y \leq 12 \)[/tex]. So, we have the point [tex]\( (0, 12) \)[/tex].
- When [tex]\( y = 0 \)[/tex], [tex]\( 4x \leq 12 \implies x \leq 3 \)[/tex]. So, we have the point [tex]\( (3, 0) \)[/tex].
Now, let's find the point of intersection of the two lines:
- [tex]\(x + 2y = 10\)[/tex]
- [tex]\(4x + y = 12\)[/tex]
Solve these equations simultaneously:
Multiply the first equation by 4:
[tex]\(4x + 8y = 40\)[/tex]
Now subtract the second equation from this result:
[tex]\((4x + 8y) - (4x + y) = 40 - 12\)[/tex]
[tex]\(7y = 28\)[/tex]
[tex]\( y = 4 \)[/tex]
Substitute [tex]\(y = 4\)[/tex] back into the first equation:
[tex]\(x + 2*4 = 10\)[/tex]
[tex]\(x + 8 = 10\)[/tex]
[tex]\(x = 2\)[/tex]
So, the point of intersection is [tex]\( (2, 4) \)[/tex].
Step 2: Identify the Feasible Region
--------------------------------------
Plotting the lines on a graph and identifying the region that satisfies all constraints is crucial. The feasible region will be bounded by the lines and the axes.
Vertices of the feasible region:
- [tex]\( (0,5) \)[/tex]
- [tex]\( (10, 0) \)[/tex]
- [tex]\( (3, 0) \)[/tex]
- [tex]\( (0, 12) \)[/tex] (but it is beyond the bound defined by [tex]\(4x+y \leq 12\)[/tex])
- [tex]\( (2, 4) \)[/tex] (intersection of [tex]\( x + 2y = 10 \)[/tex] and [tex]\( 4x + y = 12 \)[/tex])
Step 3: Evaluate the Objective Function at Each Vertex
----------------------------------------------------------
Objective function: [tex]\(P = 7x + 9y\)[/tex]
Calculate [tex]\(P\)[/tex] at each vertex:
- At [tex]\( (0, 5) \)[/tex]:
[tex]\( P = 7(0) + 9(5) = 45 \)[/tex]
- At [tex]\( (10, 0) \)[/tex]:
[tex]\( P = 7(10) + 9(0) = 70 \)[/tex]
- At [tex]\( (3, 0) \)[/tex]:
[tex]\( P = 7(3) + 9(0) = 21 \)[/tex]
- At [tex]\( (2, 4) \)[/tex]:
[tex]\( P = 7(2) + 9(4) = 14 + 36 = 50 \)[/tex]
Step 4: Determine the Maximum Value
-------------------------------------
By evaluating the objective function at each vertex, we observe that the maximum value is [tex]\(P = 50\)[/tex] at point [tex]\( (2, 4) \)[/tex].
Thus,
- [tex]\( x = 2 \)[/tex]
- [tex]\( y = 4 \)[/tex]
- Maximum value [tex]\( P = 50 \)[/tex]
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