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Q6:
Solve the following Linear Programming Problems graphically:
6. Minimise $Z = x + 2y$ subject to $2x + y \geq 3$, $x + 2y \geq 6$, $x, y \geq 0$. Show that the minimum of $Z$ occurs at more than two points.
Solution :
Step 1: Formalization of Constraints
The objective function to be minimized is $Z = x + 2y$. The constraints are defined as follows:
- Constraint 1: $2x + y \geq 3$
- Constraint 2: $x + 2y \geq 6$
- Non-negativity constraints: $x \geq 0, y \geq 0$
Step 2: Determination of Boundary Lines
To plot the feasible region, we convert the inequalities into equations:
- Line $L_1$: $2x + y = 3$.
- If $x = 0$, $y = 3$. Point: $(0, 3)$
- If $y = 0$, $x = 1.5$. Point: $(1.5, 0)$
- Line $L_2$: $x + 2y = 6$.
- If $x = 0$, $y = 3$. Point: $(0, 3)$
- If $y = 0$, $x = 6$. Point: $(6, 0)$
Step 3: Identification of Intersection Points and Corner Points
Solving the system of equations $2x + y = 3$ and $x + 2y = 6$ simultaneously:
Multiply the first equation by 2: $4x + 2y = 6$. Subtracting the second equation ($x + 2y = 6$):
$(4x - x) = 6 - 6 \implies 3x = 0 \implies x = 0$. Substituting $x=0$ into $2x + y = 3$ yields $y = 3$.
The region is bounded by the axes and these lines. Testing $(1, 2)$ shows that the shaded area lies to the right/above the lines.
Step 4: Evaluating the Objective Function at Corner Points
Since the region is unbounded in the first quadrant, we check the corner points identified by the boundary lines:
| Corner Point $(x, y)$ | $Z = x + 2y$ |
|---|---|
| $(0, 3)$ | $0 + 2(3) = 6$ |
| $(6, 0)$ | $6 + 2(0) = 6$ |
Step 5: Analysis of Multi-point Optimality
At both $(0, 3)$ and $(6, 0)$, $Z = 6$. We check if any point on the line segment connecting $(0, 3)$ and $(6, 0)$ provides a value less than 6. The line segment lies on $x + 2y = 6$. For any point $(x, y)$ on this segment, $x + 2y = 6$. Thus, $Z = 6$ for all points on the segment $x + 2y = 6$ between $x=0$ and $x=6$.
Since the problem asks for the minimum, and any point on the segment gives $Z=6$ (which is consistent with the constraint $x + 2y \geq 6$), the minimum occurs at infinitely many points along this boundary.
Final Solution: The minimum value of $Z$ is $6$, which occurs at all points on the line segment joining $(0, 3)$ and $(6, 0)$.
More Questions from Class 12 Mathematics Linear Programming EXERCISE 12.1
- Q1: Solve the following Linear Programming Problems graphically: 1. Maximise $Z = 3x + 4y$ subject to the constraints : $x + y \leq 4$, $x \geq 0$, $y \geq 0$.
- Q2: Solve the following Linear Programming Problems graphically: 2. Minimise $Z = – 3x + 4y$ subject to $x + 2y \leq 8$, $3x + 2y \leq 12$, $x \geq 0$, $y \geq 0$.
- Q3: Solve the following Linear Programming Problems graphically: 3. Maximise $Z = 5x + 3y$ subject to $3x + 5y \leq 15$, $5x + 2y \leq 10$, $x \geq 0$, $y \geq 0$.
- Q4: Solve the following Linear Programming Problems graphically: 4. Minimise $Z = 3x + 5y$ such that $x + 3y \geq 3$, $x + y \geq 2$, $x, y \geq 0$.
- Q5: Solve the following Linear Programming Problems graphically: 5. Maximise $Z = 3x + 2y$ subject to $x + 2y \leq 10$, $3x + y \leq 15$, $x, y \geq 0$.
- Q7: Solve the following Linear Programming Problems graphically: 7. Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y \leq 120$, $x + y \geq 60$, $x – 2y \geq 0$, $x, y \geq 0$.
- Q8: Solve the following Linear Programming Problems graphically: 8. Minimise and Maximise $Z = x + 2y$ subject to $x + 2y \geq 100$, $2x – y \leq 0$, $2x + y \leq 200$; $x, y \geq 0$.
- Q9: Solve the following Linear Programming Problems graphically: 9. Maximise $Z = – x + 2y$, subject to the constraints: $x \geq 3$, $x + y \geq 5$, $x + 2y \geq 6$, $y \geq 0$.
- Q10: Solve the following Linear Programming Problems graphically: 10. Maximise $Z = x + y$, subject to $x – y \leq –1$, $–x + y \leq 0$, $x, y \geq 0$.
CBSE Solutions for Class 12 Mathematics Linear Programming
Chapters in CBSE - Class 12 Mathematics
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