1. **State the problem:** We want to maximize the objective function $$Z = 2x_1 + 3x_2$$ subject to the constraints: $$4x_1 + 3x_2 \leq 18$$ $$5x_1 + 2x_2 \leq 19$$ $$x_1, x_2 \geq 0$$ 2. **Formulate the dual problem:** The primal is a maximization with \( \leq \) constraints, so the dual is a minimization problem: Let dual variables be \(y_1, y_2 \geq 0\) corresponding to the two constraints. Dual objective: $$\min W = 18y_1 + 19y_2$$ Dual constraints: $$4y_1 + 5y_2 \geq 2$$ $$3y_1 + 2y_2 \geq 3$$ $$y_1, y_2 \geq 0$$ 3. **Solve the primal problem graphically or by corner points:** - Constraint 1: $$4x_1 + 3x_2 = 18$$ - Constraint 2: $$5x_1 + 2x_2 = 19$$ Find intersection points: - When $$x_1=0$$, from constraint 1: $$x_2=6$$ - When $$x_2=0$$, from constraint 1: $$x_1=4.5$$ - When $$x_1=0$$, from constraint 2: $$x_2=9.5$$ - When $$x_2=0$$, from constraint 2: $$x_1=3.8$$ Find intersection of constraints: Solve system: $$4x_1 + 3x_2 = 18$$ $$5x_1 + 2x_2 = 19$$ Multiply first by 2 and second by 3: $$8x_1 + 6x_2 = 36$$ $$15x_1 + 6x_2 = 57$$ Subtract: $$7x_1 = 21 \Rightarrow x_1 = 3$$ Substitute back: $$4(3) + 3x_2 = 18 \Rightarrow 12 + 3x_2 = 18 \Rightarrow 3x_2 = 6 \Rightarrow x_2 = 2$$ 4. **Evaluate objective at corner points:** - At (0,0): $$Z=0$$ - At (0,6): $$Z=2(0)+3(6)=18$$ - At (4.5,0): $$Z=2(4.5)+3(0)=9$$ - At (3,2): $$Z=2(3)+3(2)=6+6=12$$ - At (3.8,0): $$Z=2(3.8)+3(0)=7.6$$ Maximum value is $$Z=18$$ at $$x_1=0, x_2=6$$. 5. **Solve the dual problem:** Check constraints: $$4y_1 + 5y_2 \geq 2$$ $$3y_1 + 2y_2 \geq 3$$ Try to find \(y_1, y_2\) minimizing $$18y_1 + 19y_2$$. Solve system as equalities: $$4y_1 + 5y_2 = 2$$ $$3y_1 + 2y_2 = 3$$ Multiply second by 5 and first by 2: $$8y_1 + 10y_2 = 4$$ $$15y_1 + 10y_2 = 15$$ Subtract: $$7y_1 = 11 \Rightarrow y_1 = \frac{11}{7} \approx 1.5714$$ Substitute back: $$3(1.5714) + 2y_2 = 3 \Rightarrow 4.7142 + 2y_2 = 3 \Rightarrow 2y_2 = -1.7142 \Rightarrow y_2 = -0.8571$$ Since \(y_2\) must be \(\geq 0\), this is not feasible. Try \(y_2=0\): $$4y_1 \geq 2 \Rightarrow y_1 \geq 0.5$$ $$3y_1 \geq 3 \Rightarrow y_1 \geq 1$$ So \(y_1 \geq 1\). Objective: $$18y_1 + 19(0) = 18y_1 \geq 18$$ Minimum at \(y_1=1, y_2=0\) with value 18. 6. **Final answers:** - Primal optimal solution: $$x_1=0, x_2=6$$ with $$Z_{max} = 18$$ - Dual optimal solution: $$y_1=1, y_2=0$$ with $$W_{min} = 18$$ This confirms strong duality: primal max = dual min = 18.