Answer :
To solve the given linear programming problem using graphical methods, we follow these steps:
1. Problem Formulation:
Maximize [tex]\( Z = 15X_1 - 10X_2 \)[/tex]
Subject to:
[tex]\[ \begin{aligned} 4X_1 + 6X_2 & \leq 360 \quad \text{(Constraint 1)}\\ 3X_1 & \leq 180 \quad \text{(Constraint 2)}\\ 5X_2 & \leq 280 \quad \text{(Constraint 3)}\\ X_1, X_2 & \geq 0 \quad \text{(Non-negativity constraints)} \end{aligned} \][/tex]
2. Plot the Constraints:
- For [tex]\( 4X_1 + 6X_2 \leq 360 \)[/tex]:
First, plot the line [tex]\( 4X_1 + 6X_2 = 360 \)[/tex].
To find the intercepts:
- When [tex]\( X_1 = 0 \)[/tex]: [tex]\( 6X_2 = 360 \implies X_2 = 60 \)[/tex]
- When [tex]\( X_2 = 0 \)[/tex]: [tex]\( 4X_1 = 360 \implies X_1 = 90 \)[/tex]
So, this line passes through points [tex]\( (0, 60) \)[/tex] and [tex]\( (90, 0) \)[/tex].
- For [tex]\( 3X_1 \leq 180 \)[/tex]:
Plot the line [tex]\( 3X_1 = 180 \)[/tex] which simplifies to [tex]\( X_1 = 60 \)[/tex].
This line is vertical at [tex]\( X_1 = 60 \)[/tex].
- For [tex]\( 5X_2 \leq 280 \)[/tex]:
Plot the line [tex]\( 5X_2 = 280 \)[/tex].
To find the intercepts:
- When [tex]\( X_1 = 0 \)[/tex]: [tex]\( 5X_2 = 280 \implies X_2 = 56 \)[/tex]
- This line is horizontal at [tex]\( X_2 = 56 \)[/tex].
3. Identify the Feasible Region:
The feasible region is bounded by the lines plotted above and is in the first quadrant since [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex] are non-negative.
- [tex]\( X_1 = 60 \)[/tex]
- [tex]\( X_2 = 56 \)[/tex]
- [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]
4. Determine the Corner Points:
The feasible region is defined by the intersection of the constraints. The key points of intersection (corner points) include:
- Intersection of [tex]\( 4X_1 + 6X_2 = 360 \)[/tex] and [tex]\( 3X_1 = 180 \)[/tex]:
- Substitute [tex]\( X_1 = 60 \)[/tex] into [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]:
[tex]\[ 4(60) + 6X_2 = 360 \][/tex]
[tex]\[ 240 + 6X_2 = 360 \][/tex]
[tex]\[ 6X_2 = 120 \implies X_2 = 20 \][/tex]
- So, intersection point is [tex]\( (60, 20) \)[/tex]
- Intersection of [tex]\( 4X_1 + 6X_2 = 360 \)[/tex] and [tex]\( 5X_2 = 280 \)[/tex]:
- Substitute [tex]\( X_2 = 56 \)[/tex] into [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]:
[tex]\[ 4X_1 + 6(56) = 360 \][/tex]
[tex]\[ 4X_1 + 336 = 360 \][/tex]
[tex]\[ 4X_1 = 24 \implies X_1 = 6 \][/tex]
- So, intersection point is [tex]\( (6, 56) \)[/tex]
- Intersection of [tex]\( X_1 = 60 \)[/tex] and [tex]\( X_2 = 0 \)[/tex]:
- Point is [tex]\( (60, 0) \)[/tex]
- Intersection of [tex]\( X_1 = 0 \)[/tex] and [tex]\( X_2 = 56 \)[/tex]:
- Point is [tex]\( (0, 56) \)[/tex]
5. Evaluate [tex]\( Z \)[/tex] at Each Corner Point:
[tex]\[ \begin{aligned} &\text{At } (60, 20): & Z &= 15(60) - 10(20) = 900 - 200 = 700\\ &\text{At } (6, 56): & Z &= 15(6) - 10(56) = 90 - 560 = -470\\ &\text{At } (60, 0): & Z &= 15(60) - 10(0) = 900\\ &\text{At } (0, 56): & Z &= 15(0) - 10(56) = 0 - 560 = -560\\ \end{aligned} \][/tex]
6. Conclusion:
The maximum value of [tex]\( Z \)[/tex] occurs at the point [tex]\( (60, 0) \)[/tex] with:
[tex]\[ \begin{aligned} X_1 & = 60\\ X_2 & = 0\\ Z & = 900 \end{aligned} \][/tex]
Hence, the optimal solution is [tex]\( X_1 = 60 \)[/tex], [tex]\( X_2 = 0 \)[/tex] and the maximum value of [tex]\( Z \)[/tex] is 900.
1. Problem Formulation:
Maximize [tex]\( Z = 15X_1 - 10X_2 \)[/tex]
Subject to:
[tex]\[ \begin{aligned} 4X_1 + 6X_2 & \leq 360 \quad \text{(Constraint 1)}\\ 3X_1 & \leq 180 \quad \text{(Constraint 2)}\\ 5X_2 & \leq 280 \quad \text{(Constraint 3)}\\ X_1, X_2 & \geq 0 \quad \text{(Non-negativity constraints)} \end{aligned} \][/tex]
2. Plot the Constraints:
- For [tex]\( 4X_1 + 6X_2 \leq 360 \)[/tex]:
First, plot the line [tex]\( 4X_1 + 6X_2 = 360 \)[/tex].
To find the intercepts:
- When [tex]\( X_1 = 0 \)[/tex]: [tex]\( 6X_2 = 360 \implies X_2 = 60 \)[/tex]
- When [tex]\( X_2 = 0 \)[/tex]: [tex]\( 4X_1 = 360 \implies X_1 = 90 \)[/tex]
So, this line passes through points [tex]\( (0, 60) \)[/tex] and [tex]\( (90, 0) \)[/tex].
- For [tex]\( 3X_1 \leq 180 \)[/tex]:
Plot the line [tex]\( 3X_1 = 180 \)[/tex] which simplifies to [tex]\( X_1 = 60 \)[/tex].
This line is vertical at [tex]\( X_1 = 60 \)[/tex].
- For [tex]\( 5X_2 \leq 280 \)[/tex]:
Plot the line [tex]\( 5X_2 = 280 \)[/tex].
To find the intercepts:
- When [tex]\( X_1 = 0 \)[/tex]: [tex]\( 5X_2 = 280 \implies X_2 = 56 \)[/tex]
- This line is horizontal at [tex]\( X_2 = 56 \)[/tex].
3. Identify the Feasible Region:
The feasible region is bounded by the lines plotted above and is in the first quadrant since [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex] are non-negative.
- [tex]\( X_1 = 60 \)[/tex]
- [tex]\( X_2 = 56 \)[/tex]
- [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]
4. Determine the Corner Points:
The feasible region is defined by the intersection of the constraints. The key points of intersection (corner points) include:
- Intersection of [tex]\( 4X_1 + 6X_2 = 360 \)[/tex] and [tex]\( 3X_1 = 180 \)[/tex]:
- Substitute [tex]\( X_1 = 60 \)[/tex] into [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]:
[tex]\[ 4(60) + 6X_2 = 360 \][/tex]
[tex]\[ 240 + 6X_2 = 360 \][/tex]
[tex]\[ 6X_2 = 120 \implies X_2 = 20 \][/tex]
- So, intersection point is [tex]\( (60, 20) \)[/tex]
- Intersection of [tex]\( 4X_1 + 6X_2 = 360 \)[/tex] and [tex]\( 5X_2 = 280 \)[/tex]:
- Substitute [tex]\( X_2 = 56 \)[/tex] into [tex]\( 4X_1 + 6X_2 = 360 \)[/tex]:
[tex]\[ 4X_1 + 6(56) = 360 \][/tex]
[tex]\[ 4X_1 + 336 = 360 \][/tex]
[tex]\[ 4X_1 = 24 \implies X_1 = 6 \][/tex]
- So, intersection point is [tex]\( (6, 56) \)[/tex]
- Intersection of [tex]\( X_1 = 60 \)[/tex] and [tex]\( X_2 = 0 \)[/tex]:
- Point is [tex]\( (60, 0) \)[/tex]
- Intersection of [tex]\( X_1 = 0 \)[/tex] and [tex]\( X_2 = 56 \)[/tex]:
- Point is [tex]\( (0, 56) \)[/tex]
5. Evaluate [tex]\( Z \)[/tex] at Each Corner Point:
[tex]\[ \begin{aligned} &\text{At } (60, 20): & Z &= 15(60) - 10(20) = 900 - 200 = 700\\ &\text{At } (6, 56): & Z &= 15(6) - 10(56) = 90 - 560 = -470\\ &\text{At } (60, 0): & Z &= 15(60) - 10(0) = 900\\ &\text{At } (0, 56): & Z &= 15(0) - 10(56) = 0 - 560 = -560\\ \end{aligned} \][/tex]
6. Conclusion:
The maximum value of [tex]\( Z \)[/tex] occurs at the point [tex]\( (60, 0) \)[/tex] with:
[tex]\[ \begin{aligned} X_1 & = 60\\ X_2 & = 0\\ Z & = 900 \end{aligned} \][/tex]
Hence, the optimal solution is [tex]\( X_1 = 60 \)[/tex], [tex]\( X_2 = 0 \)[/tex] and the maximum value of [tex]\( Z \)[/tex] is 900.
