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 \( (3,3) \).
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.