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Q9:
Solve the following Linear Programming Problems graphically:
9. Maximise $Z = – x + 2y$, subject to the constraints: $x \geq 3$, $x + y \geq 5$, $x + 2y \geq 6$, $y \geq 0$.
Solution :
Step 1: Formal Representation of Constraints
The objective function to be maximized is $Z = -x + 2y$. The constraints are defined as follows:
- $L_1: x \geq 3$
- $L_2: x + y \geq 5$
- $L_3: x + 2y \geq 6$
- $L_4: y \geq 0$
Step 2: Identification of Boundary Lines and Intersections
To identify the feasible region, we convert inequalities into equations to find the boundary lines:
| Constraint | Equation | Intercept Points |
|---|---|---|
| $L_1$ | $x = 3$ | Vertical line at $x = 3$ |
| $L_2$ | $x + y = 5$ | $(5, 0)$ and $(0, 5)$ |
| $L_3$ | $x + 2y = 6$ | $(6, 0)$ and $(0, 3)$ |
Determining Intersection Points:
- Intersection of $L_1 (x=3)$ and $L_2 (x+y=5)$: $3+y=5 \implies y=2$. Point: $(3, 2)$.
- Intersection of $L_1 (x=3)$ and $L_3 (x+2y=6)$: $3+2y=6 \implies y=1.5$. Point: $(3, 1.5)$.
- Intersection of $L_2 (x+y=5)$ and $L_3 (x+2y=6)$: Subtracting eqns: $(x+2y)-(x+y) = 6-5 \implies y=1$. Substituting $y=1$ into $x+y=5 \implies x=4$. Point: $(4, 1)$.
Step 3: Geometric Representation of the Feasible Region
Step 4: Evaluating the Corner Points
The feasible region is unbounded, extending toward infinity in the positive direction of $x$ and $y$. However, we test the vertices of the boundary:
| Corner Point $(x, y)$ | $Z = -x + 2y$ |
|---|---|
| $(3, 2)$ | $-3 + 2(2) = 1$ |
| $(4, 1)$ | $-4 + 2(1) = -2$ |
| $(6, 0)$ | $-6 + 2(0) = -6$ |
Step 5: Analysis of Unbounded Region
Per the [Corner Point Theorem for unbounded regions], we check if any point in the feasible region yields a value of $Z$ greater than the maximum found at the vertices. For $Z = -x + 2y$, as $x \to \infty$ and $y \to \infty$ along the constraint $x+y=5$, $Z$ can increase indefinitely. However, check the constraint $Z > 1$: $-x + 2y > 1$. This region overlaps with the feasible region. Therefore, the function $Z = -x + 2y$ has no finite maximum value.
Final Solution: The Linear Programming Problem has no finite maximum value because the feasible region is unbounded and the objective function $Z$ can increase indefinitely.
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 $x + 2y \leq 10$, $3x + y \leq 15$, $x, y \geq 0$.
- 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.
- 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 $x + 2y \geq 100$, $2x – y \leq 0$, $2x + y \leq 200$; $x, y \geq 0$.
- 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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