Answer :
To find the minimum value of the objective function [tex]\( C = 10x + 26y \)[/tex] subject to the given constraints:
[tex]\[ \begin{array}{l} x + y \leq 6 \\ 5x + y \geq 10 \\ x + 5y \geq 14 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
we will follow the steps of linear programming:
1. Formulate the problem:
The objective is to minimize [tex]\( C = 10x + 26y \)[/tex].
2. Identify the constraints:
[tex]\[ \begin{array}{l} x + y \leq 6 \quad \text{(Constraint 1)} \\ 5x + y \geq 10 \quad \text{(Constraint 2)} \\ x + 5y \geq 14 \quad \text{(Constraint 3)} \\ x \geq 0 \quad \text{(Non-negativity)} \\ y \geq 0 \quad \text{(Non-negativity)} \end{array} \][/tex]
3. Find the feasible region:
- Constraint 1: [tex]\( x + y = 6 \)[/tex]
- Constraint 2: [tex]\( 5x + y = 10 \)[/tex]
- Constraint 3: [tex]\( x + 5y = 14 \)[/tex]
After transforming these inequalities into equations, these constraints define half-planes. The feasible region is the intersection of these half-planes along with the non-negativity constraints.
4. Determine the corner points:
Identify the intersections of the constraints in the feasible region. Solving these intersections systematically:
- Intersection of [tex]\( x + y = 6 \)[/tex] and [tex]\( 5x + y = 10 \)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ 5x + y = 10 \end{cases} \][/tex]
Subtract the first equation from the second equation:
[tex]\[ 4x = 4 \implies x = 1 \\ y = 6 - x = 5 \][/tex]
Intersection point: [tex]\( (1, 5) \)[/tex]
- Intersection of [tex]\( x + y = 6 \)[/tex] and [tex]\( x + 5y = 14 \)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ x + 5y = 14 \end{cases} \][/tex]
Subtract the first equation from the second equation:
[tex]\[ 4y = 8 \implies y = 2 \\ x = 6 - y = 4 \][/tex]
Intersection point: [tex]\( (4, 2) \)[/tex]
- Intersection of [tex]\( 5x + y = 10 \)[/tex] and [tex]\( x + 5y = 14 \)[/tex]:
[tex]\[ \begin{cases} 5x + y = 10 \\ x + 5y = 14 \end{cases} \][/tex]
Multiply the first equation by 5:
[tex]\[ 25x + 5y = 50 \\ x + 5y = 14 \][/tex]
Subtract the second equation from this:
[tex]\[ 24x = 36 \implies x = 1.5 \\ y = (10 - 5x) = 2.5 \][/tex]
Intersection point: [tex]\( (1.5, 2.5) \)[/tex]
5. Evaluate the objective function at each corner point:
[tex]\[ C = 10x + 26y \][/tex]
- At [tex]\( (1, 5) \)[/tex]:
[tex]\[ C = 10(1) + 26(5) = 10 + 130 = 140 \][/tex]
- At [tex]\( (4, 2) \)[/tex]:
[tex]\[ C = 10(4) + 26(2) = 40 + 52 = 92 \][/tex]
- At [tex]\( (1.5, 2.5) \)[/tex]:
[tex]\[ C = 10(1.5) + 26(2.5) = 15 + 65 = 80 \][/tex]
6. Conclusion:
The minimum value of [tex]\( C = 10x + 26y \)[/tex] subject to the given constraints is found at the point [tex]\( (1.5, 2.5) \)[/tex].
Therefore, the minimum value of [tex]\( C \)[/tex] is:
[tex]\[ \boxed{80} \][/tex]
[tex]\[ \begin{array}{l} x + y \leq 6 \\ 5x + y \geq 10 \\ x + 5y \geq 14 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
we will follow the steps of linear programming:
1. Formulate the problem:
The objective is to minimize [tex]\( C = 10x + 26y \)[/tex].
2. Identify the constraints:
[tex]\[ \begin{array}{l} x + y \leq 6 \quad \text{(Constraint 1)} \\ 5x + y \geq 10 \quad \text{(Constraint 2)} \\ x + 5y \geq 14 \quad \text{(Constraint 3)} \\ x \geq 0 \quad \text{(Non-negativity)} \\ y \geq 0 \quad \text{(Non-negativity)} \end{array} \][/tex]
3. Find the feasible region:
- Constraint 1: [tex]\( x + y = 6 \)[/tex]
- Constraint 2: [tex]\( 5x + y = 10 \)[/tex]
- Constraint 3: [tex]\( x + 5y = 14 \)[/tex]
After transforming these inequalities into equations, these constraints define half-planes. The feasible region is the intersection of these half-planes along with the non-negativity constraints.
4. Determine the corner points:
Identify the intersections of the constraints in the feasible region. Solving these intersections systematically:
- Intersection of [tex]\( x + y = 6 \)[/tex] and [tex]\( 5x + y = 10 \)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ 5x + y = 10 \end{cases} \][/tex]
Subtract the first equation from the second equation:
[tex]\[ 4x = 4 \implies x = 1 \\ y = 6 - x = 5 \][/tex]
Intersection point: [tex]\( (1, 5) \)[/tex]
- Intersection of [tex]\( x + y = 6 \)[/tex] and [tex]\( x + 5y = 14 \)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ x + 5y = 14 \end{cases} \][/tex]
Subtract the first equation from the second equation:
[tex]\[ 4y = 8 \implies y = 2 \\ x = 6 - y = 4 \][/tex]
Intersection point: [tex]\( (4, 2) \)[/tex]
- Intersection of [tex]\( 5x + y = 10 \)[/tex] and [tex]\( x + 5y = 14 \)[/tex]:
[tex]\[ \begin{cases} 5x + y = 10 \\ x + 5y = 14 \end{cases} \][/tex]
Multiply the first equation by 5:
[tex]\[ 25x + 5y = 50 \\ x + 5y = 14 \][/tex]
Subtract the second equation from this:
[tex]\[ 24x = 36 \implies x = 1.5 \\ y = (10 - 5x) = 2.5 \][/tex]
Intersection point: [tex]\( (1.5, 2.5) \)[/tex]
5. Evaluate the objective function at each corner point:
[tex]\[ C = 10x + 26y \][/tex]
- At [tex]\( (1, 5) \)[/tex]:
[tex]\[ C = 10(1) + 26(5) = 10 + 130 = 140 \][/tex]
- At [tex]\( (4, 2) \)[/tex]:
[tex]\[ C = 10(4) + 26(2) = 40 + 52 = 92 \][/tex]
- At [tex]\( (1.5, 2.5) \)[/tex]:
[tex]\[ C = 10(1.5) + 26(2.5) = 15 + 65 = 80 \][/tex]
6. Conclusion:
The minimum value of [tex]\( C = 10x + 26y \)[/tex] subject to the given constraints is found at the point [tex]\( (1.5, 2.5) \)[/tex].
Therefore, the minimum value of [tex]\( C \)[/tex] is:
[tex]\[ \boxed{80} \][/tex]
