Question:

Solve the following linear programming problem graphically: Maximise \[ Z = 2x + 3y \] Subject to the constraints: \[ x + y \leq 6 \] \[ x \geq 2 \] \[ y \leq 3 \] \[ x, y \geq 0 \]

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To solve a linear programming problem graphically, plot the constraints, identify the feasible region, determine the corner points, and evaluate the objective function at each point.
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Solution and Explanation

Step 1: Graphical Representation of Constraints. To solve the problem graphically, we plot the constraint equations: 1. \( x + y = 6 \) (line passing through (6,0) and (0,6)) 2. \( x = 2 \) (vertical line at \( x = 2 \)) 3. \( y = 3 \) (horizontal line at \( y = 3 \)) 4. \( x, y \geq 0 \) (first quadrant restriction) The feasible region is the intersection of these constraints. 

Step 2: Identifying Corner Points of Feasible Region. From the graph, the common feasible region forms a bounded polygon. The corner points of this region are: \[ A(2,0), B(2,3), C(3,3), D(6,0) \] 

Step 3: Compute Objective Function at Corner Points. Evaluating \( Z = 2x + 3y \) at each corner: \[ Z(A) = 2(2) + 3(0) = 4 \] \[ Z(B) = 2(2) + 3(3) = 13 \] \[ Z(C) = 2(3) + 3(3) = 15 \] \[ Z(D) = 2(6) + 3(0) = 12 \] 

Step 4: Determine Maximum Value. The maximum value occurs at point \( C(3,3) \) with: \[ Z_{\max} = 15 \] 
Z = x + 3y

Conclusion: The maximum value of \( Z \) is \( 15 \) at \( (3,3) \).
Z = x + 3y

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