Class CBSE Class 12 Mathematics Linear Programming Q #1455
COMPETENCY BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
Consider the Linear Programming Problem, where the objective function $Z=(x+4y)$ needs to be minimized subject to constraints $2x+y\ge1000$, $x+2y\ge800$, $x,y\ge0$. Draw a neat graph of the feasible region and find the minimum value of Z.

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Detailed Solution

Step 1: Graph the Constraints

First, we need to graph the inequalities $2x+y\ge1000$ and $x+2y\ge800$, along with the non-negativity constraints $x,y\ge0$.

Step 2: Convert Inequalities to Equations

Convert the inequalities to equations to find the boundary lines: $2x+y=1000$ and $x+2y=800$.

Step 3: Find Intercepts

For $2x+y=1000$: If $x=0$, then $y=1000$. Point: $(0, 1000)$ If $y=0$, then $2x=1000 \implies x=500$. Point: $(500, 0)$ For $x+2y=800$: If $x=0$, then $2y=800 \implies y=400$. Point: $(0, 400)$ If $y=0$, then $x=800$. Point: $(800, 0)$

Step 4: Plot the Lines

Plot the lines $2x+y=1000$ and $x+2y=800$ on the graph. Since the inequalities are $\ge$, shade the region above each line.

Step 5: Identify the Feasible Region

The feasible region is the intersection of the shaded regions, which is unbounded. The corner points of the feasible region are the intersection of the lines and the axes.

Step 6: Find the Intersection Point

Solve the system of equations: $2x+y=1000$ $x+2y=800$ Multiply the second equation by 2: $2x+4y=1600$. Subtract the first equation from the modified second equation: $(2x+4y) - (2x+y) = 1600 - 1000$ $3y = 600$ $y = 200$ Substitute $y=200$ into $x+2y=800$: $x+2(200)=800$ $x+400=800$ $x=400$ The intersection point is $(400, 200)$.

Step 7: Identify Corner Points

The corner points of the feasible region are $(0, 1000)$, $(400, 200)$, and $(800, 0)$.

Step 8: Evaluate the Objective Function

Evaluate the objective function $Z=x+4y$ at each corner point: At $(0, 1000)$: $Z = 0 + 4(1000) = 4000$ At $(400, 200)$: $Z = 400 + 4(200) = 400 + 800 = 1200$ At $(800, 0)$: $Z = 800 + 4(0) = 800$

Step 9: Determine the Minimum Value

The minimum value of $Z$ is $800$ at the point $(800, 0)$.

Final Answer: 800

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the steps to solve a linear programming problem, including graphing constraints, identifying the feasible region, and evaluating the objective function at corner points.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific algorithm or method (linear programming) to find the minimum value of the objective function. This involves a series of steps, such as graphing, finding intersection points, and evaluating the objective function.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply the concepts of linear programming to solve a problem, including graphing and optimization.