CBSE Class 12 Mathematics Linear Programming Q #1798
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
Solve the following Linear Programming Problem graphically: Maximise $Z=200x+120y$ subject to the constraints $x+y\le 300$, $3x+y\le 600$, $x-y\ge -100$, $x,y\ge 0$
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Detailed Solution

Step 1: Identify Constraints and Boundary Lines

The constraints are: $$x + y = 300$$ $$3x + y = 600$$ $$x - y = -100$$ $$x \ge 0, y \ge 0$$

Step 2: Find Intersection Points

For $x+y=300$, points are $(300,0)$ and $(0,300)$. For $3x+y=600$, points are $(200,0)$ and $(0,600)$. For $x-y=-100$, points are $(0,100)$ and $(-100,0)$.

Step 3: Determine Feasible Region and Vertices

The intersection of the half-planes defined by the constraints yields the vertices of the feasible region: 1. $(0,0)$ 2. $(0,100)$ (Intersection of $x=0$ and $x-y=-100$) 3. $(150, 150)$ (Intersection of $x+y=300$ and $x-y=-100$) 4. $(200, 0)$ (Intersection of $3x+y=600$ and $y=0$)

Step 4: Evaluate Objective Function $Z=200x+120y$

At $(0,0): Z = 0$ At $(0,100): Z = 200(0) + 120(100) = 12,000$ At $(150,150): Z = 200(150) + 120(150) = 30,000 + 18,000 = 48,000$ At $(200,0): Z = 200(200) + 120(0) = 40,000$

Final Answer: Maximum Z = 48,000 at point (150, 150)

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to translate algebraic inequalities into a geometric representation and apply the corner point method to optimize a linear function.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of steps: graphing lines, identifying the feasible region, finding vertices, and testing the objective function.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. Linear Programming is a core application-based topic in the Class 12 curriculum, focusing on decision-making under constraints.

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