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Sagot :
Certainly, let's solve the given linear programming problem step-by-step.
### Problem Statement:
Maximize the objective function:
[tex]\[ Z = 4x_1 + 8x_2 \][/tex]
Subject to the constraints:
1. [tex]\( 2x_1 + 3x_2 \leq 48 \)[/tex]
2. [tex]\( x_1 \leq 15 \)[/tex]
3. [tex]\( x_2 \leq 10 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
### Step 1: Formulate the inequalities
We need to graph the inequalities and identify the feasible region.
1. [tex]\( 2x_1 + 3x_2 \leq 48 \)[/tex]
2. [tex]\( x_1 \leq 15 \)[/tex]
3. [tex]\( x_2 \leq 10 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
### Step 2: Identify the corner points
We need to find the points of intersection of the lines representing the constraints.
1. For [tex]\( 2x_1 + 3x_2 = 48 \)[/tex] and [tex]\( x_1 = 0 \)[/tex]:
[tex]\[ 2(0) + 3x_2 = 48 \implies x_2 = 16 \implies (0, 16) \][/tex]
2. For [tex]\( 2x_1 + 3x_2 = 48 \)[/tex] and [tex]\( x_2 = 0 \)[/tex]:
[tex]\[ 2x_1 + 3(0) = 48 \implies x_1 = 24 \implies (24, 0) \][/tex]
3. For [tex]\( x_1 = 15 \)[/tex] and [tex]\( x_1 = 15 \)[/tex]:
[tex]\[ (15, 0) \][/tex]
4. For [tex]\( x_2 = 10 \)[/tex] and [tex]\( x_2 = 10 \)[/tex]:
[tex]\[ (0, 10) \][/tex]
5. For the intersection of [tex]\( 2x_1 + 3x_2 = 48 \)[/tex], [tex]\( x_1 \leq 15 \)[/tex] and [tex]\( x_2 \leq 10 \)[/tex]:
Plug [tex]\( x_1 = 15 \)[/tex] into [tex]\( 2x_1 + 3x_2 = 48 \)[/tex]:
[tex]\[ 2(15) + 3x_2 = 48 \implies 30 + 3x_2 = 48 \implies 3x_2 = 18 \implies x_2 = 6 \implies (15, 6) \][/tex]
6. For the intersection of [tex]\( 2x_1 + 3x_2 = 48 \)[/tex], [tex]\( x_2 \leq 10 \)[/tex]:
[tex]\[ x_2 = 10 \implies 2x_1 + 3(10) = 48 \implies 2x_1 + 30 = 48 \implies 2x_1 = 18 \implies x_1 = 9 \implies (9, 10) \][/tex]
### Step 3: Evaluate the objective function at each corner point
1. At [tex]\( (0, 10) \)[/tex]:
[tex]\[ Z = 4(0) + 8(10) = 80 \][/tex]
2. At [tex]\( (9, 10) \)[/tex]:
[tex]\[ Z = 4(9) + 8(10) = 36 + 80 = 116 \][/tex]
3. At [tex]\( (15, 6) \)[/tex]:
[tex]\[ Z = 4(15) + 8(6) = 60 + 48 = 108 \][/tex]
4. At [tex]\( (15, 0) \)[/tex]:
[tex]\[ Z = 4(15) + 8(0) = 60 \][/tex]
5. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ Z = 4(0) + 8(0) = 0 \][/tex]
### Step 4: Choose the maximum value
Comparing all the Z values, the maximum value of [tex]\( Z \)[/tex] is at the point [tex]\( (9, 10) \)[/tex] with [tex]\( Z = 116 \)[/tex].
### Conclusion
The optimal solution to maximize the objective function [tex]\( Z = 4x_1 + 8x_2 \)[/tex] under the given constraints is:
[tex]\[ x_1 = 9, \quad x_2 = 10, \quad \text{and} \quad Z_{\max} = 116 \][/tex]
Thus, the optimal values are [tex]\( x_1 = 9 \)[/tex], [tex]\( x_2 = 10 \)[/tex], giving a maximum objective function value of [tex]\( Z = 116 \)[/tex].
### Problem Statement:
Maximize the objective function:
[tex]\[ Z = 4x_1 + 8x_2 \][/tex]
Subject to the constraints:
1. [tex]\( 2x_1 + 3x_2 \leq 48 \)[/tex]
2. [tex]\( x_1 \leq 15 \)[/tex]
3. [tex]\( x_2 \leq 10 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
### Step 1: Formulate the inequalities
We need to graph the inequalities and identify the feasible region.
1. [tex]\( 2x_1 + 3x_2 \leq 48 \)[/tex]
2. [tex]\( x_1 \leq 15 \)[/tex]
3. [tex]\( x_2 \leq 10 \)[/tex]
4. [tex]\( x_1 \geq 0 \)[/tex]
5. [tex]\( x_2 \geq 0 \)[/tex]
### Step 2: Identify the corner points
We need to find the points of intersection of the lines representing the constraints.
1. For [tex]\( 2x_1 + 3x_2 = 48 \)[/tex] and [tex]\( x_1 = 0 \)[/tex]:
[tex]\[ 2(0) + 3x_2 = 48 \implies x_2 = 16 \implies (0, 16) \][/tex]
2. For [tex]\( 2x_1 + 3x_2 = 48 \)[/tex] and [tex]\( x_2 = 0 \)[/tex]:
[tex]\[ 2x_1 + 3(0) = 48 \implies x_1 = 24 \implies (24, 0) \][/tex]
3. For [tex]\( x_1 = 15 \)[/tex] and [tex]\( x_1 = 15 \)[/tex]:
[tex]\[ (15, 0) \][/tex]
4. For [tex]\( x_2 = 10 \)[/tex] and [tex]\( x_2 = 10 \)[/tex]:
[tex]\[ (0, 10) \][/tex]
5. For the intersection of [tex]\( 2x_1 + 3x_2 = 48 \)[/tex], [tex]\( x_1 \leq 15 \)[/tex] and [tex]\( x_2 \leq 10 \)[/tex]:
Plug [tex]\( x_1 = 15 \)[/tex] into [tex]\( 2x_1 + 3x_2 = 48 \)[/tex]:
[tex]\[ 2(15) + 3x_2 = 48 \implies 30 + 3x_2 = 48 \implies 3x_2 = 18 \implies x_2 = 6 \implies (15, 6) \][/tex]
6. For the intersection of [tex]\( 2x_1 + 3x_2 = 48 \)[/tex], [tex]\( x_2 \leq 10 \)[/tex]:
[tex]\[ x_2 = 10 \implies 2x_1 + 3(10) = 48 \implies 2x_1 + 30 = 48 \implies 2x_1 = 18 \implies x_1 = 9 \implies (9, 10) \][/tex]
### Step 3: Evaluate the objective function at each corner point
1. At [tex]\( (0, 10) \)[/tex]:
[tex]\[ Z = 4(0) + 8(10) = 80 \][/tex]
2. At [tex]\( (9, 10) \)[/tex]:
[tex]\[ Z = 4(9) + 8(10) = 36 + 80 = 116 \][/tex]
3. At [tex]\( (15, 6) \)[/tex]:
[tex]\[ Z = 4(15) + 8(6) = 60 + 48 = 108 \][/tex]
4. At [tex]\( (15, 0) \)[/tex]:
[tex]\[ Z = 4(15) + 8(0) = 60 \][/tex]
5. At [tex]\( (0, 0) \)[/tex]:
[tex]\[ Z = 4(0) + 8(0) = 0 \][/tex]
### Step 4: Choose the maximum value
Comparing all the Z values, the maximum value of [tex]\( Z \)[/tex] is at the point [tex]\( (9, 10) \)[/tex] with [tex]\( Z = 116 \)[/tex].
### Conclusion
The optimal solution to maximize the objective function [tex]\( Z = 4x_1 + 8x_2 \)[/tex] under the given constraints is:
[tex]\[ x_1 = 9, \quad x_2 = 10, \quad \text{and} \quad Z_{\max} = 116 \][/tex]
Thus, the optimal values are [tex]\( x_1 = 9 \)[/tex], [tex]\( x_2 = 10 \)[/tex], giving a maximum objective function value of [tex]\( Z = 116 \)[/tex].
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