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Q7:
Solve the following Linear Programming Problems graphically:
7. Minimise and Maximise $Z = 5x + 10y$ subject to $x + 2y \leq 120$, $x + y \geq 60$, $x – 2y \geq 0$, $x, y \geq 0$.
Solution :
Step 1: Formal Identification of Constraints and Objective Function
We aim to optimize the objective function $Z = 5x + 10y$ subject to the following system of linear inequalities:
- $L_1: x + 2y \leq 120$
- $L_2: x + y \geq 60$
- $L_3: x - 2y \geq 0$
- Non-negativity constraints: $x \geq 0, y \geq 0$
Step 2: Conversion to Equations for Boundary Plotting
To identify the feasible region, we define the boundary lines:
| Line | Equation | Intercepts/Points |
|---|---|---|
| $L_1$ | $x + 2y = 120$ | $(120, 0), (0, 60)$ |
| $L_2$ | $x + y = 60$ | $(60, 0), (0, 60)$ |
| $L_3$ | $x - 2y = 0$ | $(0, 0), (40, 20), (60, 30)$ |
Step 3: Determination of the Feasible Region
The feasible region is the intersection of the half-planes defined by the constraints. [Per the Graphical Method for LPP], we test the origin $(0,0)$ or convenient points:
- For $x + 2y \leq 120$: Since $0 + 0 \leq 120$, the region includes the origin.
- For $x + y \geq 60$: Since $0 + 0 \ngeq 60$, the region is away from the origin.
- For $x - 2y \geq 0$: Testing $(60, 0)$, $60 - 0 \geq 0$ holds true. The region lies below the line $y = 0.5x$.
Step 4: Calculation of Corner Points
The vertices of the feasible region are found by solving the intersection of the boundary lines:
- Point A (Intersection of $L_2$ and $L_3$): $x+y=60$ and $x=2y \Rightarrow 3y=60 \Rightarrow y=20, x=40$. Point: $(40, 20)$.
- Point B (Intersection of $L_1$ and $L_3$): $x+2y=120$ and $x=2y \Rightarrow 4y=120 \Rightarrow y=30, x=60$. Point: $(60, 30)$.
- Point C (Intersection of $L_1$ and $L_2$): $x+2y=120$ and $x+y=60 \Rightarrow y=60, x=0$. Point: $(0, 60)$.
Step 5: Evaluating the Objective Function $Z = 5x + 10y$
| Corner Point | $Z = 5x + 10y$ |
|---|---|
| $(40, 20)$ | $5(40) + 10(20) = 200 + 200 = 400$ |
| $(60, 30)$ | $5(60) + 10(30) = 300 + 300 = 600$ |
| $(0, 60)$ | $5(0) + 10(60) = 0 + 600 = 600$ |
Final Solution: The minimum value of $Z$ is $400$ at point $(40, 20)$, and the maximum value of $Z$ is $600$ at all points on the segment joining $(60, 30)$ and $(0, 60)$.
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.
- 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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