Answer :
Sure, let's solve the given linear programming problem step-by-step.
### Problem Statement
We need to maximize the objective function [tex]\( P = 6x + 2y \)[/tex] subject to the constraints:
[tex]\[ \begin{array}{l} \left\{ \begin{array}{l} 4x - y \geq 1 \\ x \geq 0 \\ x \leq 5 \\ y \geq 0 \end{array} \right. \end{array} \][/tex]
### Step 1: Identify the Constraints and the Feasible Region
The constraints define the feasible region within which we need to find the maximum value of [tex]\( P \)[/tex].
1. [tex]\( 4x - y \geq 1 \)[/tex] can be rephrased for graphing purposes as [tex]\( y \leq 4x - 1 \)[/tex].
2. [tex]\( x \geq 0 \)[/tex] ensures [tex]\( x \)[/tex] is non-negative.
3. [tex]\( x \leq 5 \)[/tex] bounds [tex]\( x \)[/tex] to the left by 5.
4. [tex]\( y \geq 0 \)[/tex] ensures [tex]\( y \)[/tex] is non-negative.
### Step 2: Locate the Boundary Lines
Let's graph the boundary lines and determine the intersection points:
- For [tex]\( 4x - y \geq 1 \)[/tex]:
[tex]\[ y = 4x - 1 \][/tex]
- Graph [tex]\( x \geq 0 \)[/tex] which is the [tex]\( y \)[/tex]-axis.
- Graph [tex]\( x \leq 5 \)[/tex], which is the vertical line at [tex]\( x = 5 \)[/tex].
- Graph [tex]\( y \geq 0 \)[/tex], which is the [tex]\( x \)[/tex]-axis.
### Step 3: Identify Corner/Vertex Points
Find the points where the boundary lines intersect, as these vertices define the feasible region where the maximum value of [tex]\( P \)[/tex] can occur:
1. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -1 \quad (\text{Not feasible as } y \geq 0) \][/tex]
2. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 4(5) - 1 = 20 - 1 = 19 \][/tex]
So, this gives the point [tex]\( (5, 19) \)[/tex].
3. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 4x - 1 \implies 4x = 1 \implies x = \frac{1}{4} \][/tex]
So, this gives the point [tex]\( \left( \frac{1}{4}, 0 \right) \)[/tex].
Other feasible vertices include any intersection of the constraints:
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (5, 0) \)[/tex]
### Step 4: Evaluate the Objective Function at Each Vertex
We'll plug these vertex coordinates back into [tex]\( P = 6x + 2y \)[/tex]:
1. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ P = 6(0) + 2(0) = 0 \][/tex]
2. At [tex]\( (5, 0) \)[/tex]:
[tex]\[ P = 6(5) + 2(0) = 30 \][/tex]
3. At [tex]\( \left( \frac{1}{4}, 0 \right) \)[/tex]:
[tex]\[ P = 6 \left( \frac{1}{4} \right) + 2(0) = \frac{6}{4} = 1.5 \][/tex]
4. At [tex]\( (5, 19) \)[/tex]:
[tex]\[ P = 6(5) + 2(19) = 30 + 38 = 68 \][/tex]
### Step 5: Determine the Maximum Value
After evaluating [tex]\( P \)[/tex] at all feasible vertices, the maximum value of [tex]\( P \)[/tex] is found to be [tex]\( 68 \)[/tex] at the point [tex]\( (5, 19) \)[/tex].
### Conclusion
The maximum value of [tex]\( P = 6x + 2y \)[/tex] is:
[tex]\[ P = 68 \][/tex]
attu()+ [tex]\( x = 5 \)[/tex], and [tex]\( y = 19 \)[/tex].
### Problem Statement
We need to maximize the objective function [tex]\( P = 6x + 2y \)[/tex] subject to the constraints:
[tex]\[ \begin{array}{l} \left\{ \begin{array}{l} 4x - y \geq 1 \\ x \geq 0 \\ x \leq 5 \\ y \geq 0 \end{array} \right. \end{array} \][/tex]
### Step 1: Identify the Constraints and the Feasible Region
The constraints define the feasible region within which we need to find the maximum value of [tex]\( P \)[/tex].
1. [tex]\( 4x - y \geq 1 \)[/tex] can be rephrased for graphing purposes as [tex]\( y \leq 4x - 1 \)[/tex].
2. [tex]\( x \geq 0 \)[/tex] ensures [tex]\( x \)[/tex] is non-negative.
3. [tex]\( x \leq 5 \)[/tex] bounds [tex]\( x \)[/tex] to the left by 5.
4. [tex]\( y \geq 0 \)[/tex] ensures [tex]\( y \)[/tex] is non-negative.
### Step 2: Locate the Boundary Lines
Let's graph the boundary lines and determine the intersection points:
- For [tex]\( 4x - y \geq 1 \)[/tex]:
[tex]\[ y = 4x - 1 \][/tex]
- Graph [tex]\( x \geq 0 \)[/tex] which is the [tex]\( y \)[/tex]-axis.
- Graph [tex]\( x \leq 5 \)[/tex], which is the vertical line at [tex]\( x = 5 \)[/tex].
- Graph [tex]\( y \geq 0 \)[/tex], which is the [tex]\( x \)[/tex]-axis.
### Step 3: Identify Corner/Vertex Points
Find the points where the boundary lines intersect, as these vertices define the feasible region where the maximum value of [tex]\( P \)[/tex] can occur:
1. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -1 \quad (\text{Not feasible as } y \geq 0) \][/tex]
2. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 4(5) - 1 = 20 - 1 = 19 \][/tex]
So, this gives the point [tex]\( (5, 19) \)[/tex].
3. Intersection of [tex]\( y = 4x - 1 \)[/tex] and [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 4x - 1 \implies 4x = 1 \implies x = \frac{1}{4} \][/tex]
So, this gives the point [tex]\( \left( \frac{1}{4}, 0 \right) \)[/tex].
Other feasible vertices include any intersection of the constraints:
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (5, 0) \)[/tex]
### Step 4: Evaluate the Objective Function at Each Vertex
We'll plug these vertex coordinates back into [tex]\( P = 6x + 2y \)[/tex]:
1. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ P = 6(0) + 2(0) = 0 \][/tex]
2. At [tex]\( (5, 0) \)[/tex]:
[tex]\[ P = 6(5) + 2(0) = 30 \][/tex]
3. At [tex]\( \left( \frac{1}{4}, 0 \right) \)[/tex]:
[tex]\[ P = 6 \left( \frac{1}{4} \right) + 2(0) = \frac{6}{4} = 1.5 \][/tex]
4. At [tex]\( (5, 19) \)[/tex]:
[tex]\[ P = 6(5) + 2(19) = 30 + 38 = 68 \][/tex]
### Step 5: Determine the Maximum Value
After evaluating [tex]\( P \)[/tex] at all feasible vertices, the maximum value of [tex]\( P \)[/tex] is found to be [tex]\( 68 \)[/tex] at the point [tex]\( (5, 19) \)[/tex].
### Conclusion
The maximum value of [tex]\( P = 6x + 2y \)[/tex] is:
[tex]\[ P = 68 \][/tex]
attu()+ [tex]\( x = 5 \)[/tex], and [tex]\( y = 19 \)[/tex].
