Q1414: Solve the following Linear Programming Problem using graphic...

Class CBSE Class 12 Mathematics Linear Programming Q #1414

COMPETENCY BASED

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3 Marks 2025 AISSCE(Board Exam) SA

Solve the following Linear Programming Problem using graphical method: Maximise $Z=100x+50y$ subject to the constraints $3x+y\le600$, $x+y\le300$, $y\le x+200$, $x\ge0$, $y\ge0$.

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Detailed Solution

Step 1: Convert inequalities to equations

Convert the inequalities into equations to find the boundary lines of the feasible region:\r\n$3x + y = 600$\r\n$x + y = 300$\r\n$y = x + 200$\r\n$x = 0$\r\n$y = 0$

Step 2: Find the intersection points

Find the intersection points of these lines:\r\nIntersection of $3x + y = 600$ and $x + y = 300$:\r\nSubtract the second equation from the first:\r\n$2x = 300 \implies x = 150$\r\nSubstitute $x = 150$ into $x + y = 300$:\r\n$150 + y = 300 \implies y = 150$\r\nSo, the intersection point is $(150, 150)$.\r\n\r\nIntersection of $x + y = 300$ and $y = x + 200$:\r\nSubstitute $y = x + 200$ into $x + y = 300$:\r\n$x + (x + 200) = 300 \implies 2x = 100 \implies x = 50$\r\n$y = 50 + 200 = 250$\r\nSo, the intersection point is $(50, 250)$.\r\n\r\nIntersection of $3x + y = 600$ and $y = x + 200$:\r\nSubstitute $y = x + 200$ into $3x + y = 600$:\r\n$3x + (x + 200) = 600 \implies 4x = 400 \implies x = 100$\r\n$y = 100 + 200 = 300$\r\nSo, the intersection point is $(100, 300)$.\r\n\r\nIntersection of $3x+y=600$ and $x=0$: $(0,600)$\r\nIntersection of $x+y=300$ and $x=0$: $(0,300)$\r\nIntersection of $y=x+200$ and $x=0$: $(0,200)$\r\nIntersection of $3x+y=600$ and $y=0$: $(200,0)$\r\nIntersection of $x+y=300$ and $y=0$: $(300,0)$\r\nIntersection of $y=x+200$ and $y=0$: $(-200,0)$ but this is not in the feasible region since $x \ge 0$ and $y \ge 0$.

Step 3: Identify the feasible region

The feasible region is bounded by the constraints $3x + y \le 600$, $x + y \le 300$, $y \le x + 200$, $x \ge 0$, and $y \ge 0$. The corner points of the feasible region are $(0, 0)$, $(200, 0)$, $(150, 150)$, $(50, 250)$, and $(0, 200)$.

Step 4: Evaluate the objective function at the corner points

Evaluate $Z = 100x + 50y$ at each corner point:\r\nAt $(0, 0)$: $Z = 100(0) + 50(0) = 0$\r\nAt $(200, 0)$: $Z = 100(200) + 50(0) = 20000$\r\nAt $(150, 150)$: $Z = 100(150) + 50(150) = 15000 + 7500 = 22500$\r\nAt $(50, 250)$: $Z = 100(50) + 50(250) = 5000 + 12500 = 17500$\r\nAt $(0, 200)$: $Z = 100(0) + 50(200) = 10000$

Step 5: Determine the maximum value

The maximum value of $Z$ is $22500$ at the point $(150, 150)$.

Final Answer: Maximum $Z = 22500$ at $x = 150$ and $y = 150$<\/span>

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the steps involved in solving a linear programming problem using the graphical method.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific algorithm or method (graphical method) to solve the linear programming problem. This involves a series of steps, such as converting inequalities to equations, finding intersection points, identifying the feasible region, and evaluating the objective function at corner points.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the graphical method to solve a linear programming problem, which is a practical application of the concepts learned.