Q1302: Solve the following linear programming problem graphically: ...

Class CBSE Class 12 Mathematics Linear Programming Q #1302

COMPETENCY BASED

UNDERSTAND

3 Marks 2024 AISSCE(Board Exam) SA

Solve the following linear programming problem graphically: Maximise $z=500x+300y,$ subject to constraints $x+2y\le12$, $2x+y\le12$, $4x+5y\ge20$, $x\ge0$, $y\ge0$

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AI Tutor Explanation

Detailed Solution

Step 1: Graph the Constraints

First, we graph the inequalities to find the feasible region. $x+2y\le12$ can be written as $x+2y=12$. The points are $(12,0)$ and $(0,6)$. $2x+y\le12$ can be written as $2x+y=12$. The points are $(6,0)$ and $(0,12)$. $4x+5y\ge20$ can be written as $4x+5y=20$. The points are $(5,0)$ and $(0,4)$. $x\ge0$ and $y\ge0$ restrict the solution to the first quadrant.

Step 2: Identify the Feasible Region

The feasible region is the intersection of all the inequalities. By plotting the lines and checking the inequalities, we find the feasible region is a polygon with vertices at the intersection points of the lines.

Step 3: Find the Vertices of the Feasible Region

The vertices are the intersection points of the lines. Intersection of $x+2y=12$ and $2x+y=12$: Multiply the first equation by 2: $2x+4y=24$. Subtract the second equation: $3y=12$, so $y=4$. Then $x+2(4)=12$, so $x=4$. The intersection point is $(4,4)$. Intersection of $2x+y=12$ and $4x+5y=20$: Multiply the first equation by 5: $10x+5y=60$. Subtract the second equation: $6x=40$, so $x=\frac{20}{3}$. Then $2(\frac{20}{3})+y=12$, so $y=12-\frac{40}{3}=\frac{36-40}{3}=-\frac{4}{3}$. This point is not in the feasible region since $y$ must be non-negative. Intersection of $x+2y=12$ and $4x+5y=20$: Multiply the first equation by 4: $4x+8y=48$. Subtract the second equation: $3y=28$, so $y=\frac{28}{3}$. Then $x+2(\frac{28}{3})=12$, so $x=12-\frac{56}{3}=\frac{36-56}{3}=-\frac{20}{3}$. This point is not in the feasible region since $x$ must be non-negative. Intersection of $4x+5y=20$ and $x=0$: $4(0)+5y=20$, so $y=4$. The point is $(0,4)$. Intersection of $4x+5y=20$ and $y=0$: $4x+5(0)=20$, so $x=5$. The point is $(5,0)$. Intersection of $x+2y=12$ and $x=0$: $0+2y=12$, so $y=6$. The point is $(0,6)$. Intersection of $2x+y=12$ and $y=0$: $2x+0=12$, so $x=6$. The point is $(6,0)$. The vertices of the feasible region are $(5,0)$, $(6,0)$, $(4,4)$, and $(0,4)$.

Step 4: Evaluate the Objective Function at Each Vertex

Evaluate $z=500x+300y$ at each vertex: At $(5,0)$: $z=500(5)+300(0)=2500$. At $(6,0)$: $z=500(6)+300(0)=3000$. At $(4,4)$: $z=500(4)+300(4)=2000+1200=3200$. At $(0,4)$: $z=500(0)+300(4)=1200$. At $(0,6)$: $z=500(0)+300(6)=1800$.

Step 5: Determine the Maximum Value

The maximum value of $z$ is $3200$ at the point $(4,4)$.

Final Answer: The maximum value of $z$ is 3200 at $(4,4)$.

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because it requires the student to understand the concepts of linear programming, constraints, feasible region, and objective function to find the optimal solution.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific algorithm (graphical method) to solve the linear programming problem. This involves graphing the constraints, identifying the feasible region, finding the vertices, and evaluating the objective function at each vertex.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to apply the concepts of linear programming to solve a practical problem.