Q1354: Solve the following L.P.P. graphically: Maximise $Z=60x+40y$...

Class CBSE Class 12 Mathematics Linear Programming Q #1354

COMPETENCY BASED

UNDERSTAND

5 Marks 2024 AISSCE(Board Exam) LA

Solve the following L.P.P. graphically: Maximise $Z=60x+40y$ Subject to $x+2y\le12$, $2x+y\le12$, $4x+5y\ge20$, $x,y\ge0$

Prev Next

AI Tutor Explanation

Detailed Solution

Step 1: Convert inequalities to equations

Convert the inequalities into equations to represent the lines on the graph.

Step 2: Plot the lines

Plot the lines corresponding to the equations: $x + 2y = 12$ $2x + y = 12$ $4x + 5y = 20$ $x = 0$ $y = 0$

Step 3: Determine the feasible region

Determine the feasible region by considering the inequalities. $x + 2y \le 12$ $2x + y \le 12$ $4x + 5y \ge 20$ $x \ge 0$ $y \ge 0$ The feasible region is the area that satisfies all these inequalities.

Step 4: Find the corner points

Identify the corner points of the feasible region. These are the points where the lines intersect. 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 $x = 12 - 2(4) = 4$. So the 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 $y = 12 - 2(\frac{20}{3}) = 12 - \frac{40}{3} = \frac{36 - 40}{3} = -\frac{4}{3}$. 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 $x = 12 - 2(\frac{28}{3}) = 12 - \frac{56}{3} = \frac{36 - 56}{3} = -\frac{20}{3}$. This point is not in the feasible region since $x$ must be non-negative. Intersection of $x + 2y = 12$ and $x = 0$: $(0, 6)$ Intersection of $2x + y = 12$ and $y = 0$: $(6, 0)$ Intersection of $4x + 5y = 20$ and $x = 0$: $(0, 4)$ Intersection of $4x + 5y = 20$ and $y = 0$: $(5, 0)$ The corner points are $(0, 4)$, $(0, 6)$, $(5, 0)$, $(6, 0)$, and $(4, 4)$.

Step 5: Evaluate the objective function

Evaluate the objective function $Z = 60x + 40y$ at each corner point: At $(0, 4)$: $Z = 60(0) + 40(4) = 160$ At $(0, 6)$: $Z = 60(0) + 40(6) = 240$ At $(5, 0)$: $Z = 60(5) + 40(0) = 300$ At $(6, 0)$: $Z = 60(6) + 40(0) = 360$ At $(4, 4)$: $Z = 60(4) + 40(4) = 240 + 160 = 400$

Step 6: Determine the maximum value

The maximum value of $Z$ is $400$ at the point $(4, 4)$.

Final Answer: Maximum $Z = 400$ at $(4, 4)$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concepts of linear programming, inequalities, and graphical representation to solve the problem.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a specific procedure (graphical method) to solve a linear programming problem. This involves converting inequalities to equations, plotting lines, identifying the feasible region, finding corner points, and evaluating the objective function.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of linear programming to solve a practical problem, requiring them to interpret and analyze the given constraints and objective function.