Solve the following linear programming problem graphically:
Maximise
\[
Z = 2x + 3y
\]
Subject to the constraints:
\[
x + y \leq 6
\]
\[
x \geq 2
\]
\[
y \leq 3
\]
\[
x, y \geq 0
\]
Show Hint
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: \[ Z_{\max} = 15 \] 
Conclusion: The maximum value of \( Z \) is \( 15 \) at \( (3,3) \).
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