1. **State the problem:** Maximize the objective function $$Z = 40X_1 + 32X_2$$ subject to the constraints: $$40X_1 + 20X_2 \leq 600$$ $$4X_1 + 10X_2 \leq 100$$ $$2X_1 + 3X_2 \leq 38$$ $$X_1, X_2 \geq 0$$ 2. **Explain the method:** We will use the graphical method to find the feasible region defined by the inequalities and then evaluate the objective function at each vertex (corner point) of this region. 3. **Find intercepts for each constraint:** - For $$40X_1 + 20X_2 \leq 600$$: - If $$X_1=0$$, then $$20X_2=600 \Rightarrow X_2=30$$ - If $$X_2=0$$, then $$40X_1=600 \Rightarrow X_1=15$$ - For $$4X_1 + 10X_2 \leq 100$$: - If $$X_1=0$$, then $$10X_2=100 \Rightarrow X_2=10$$ - If $$X_2=0$$, then $$4X_1=100 \Rightarrow X_1=25$$ - For $$2X_1 + 3X_2 \leq 38$$: - If $$X_1=0$$, then $$3X_2=38 \Rightarrow X_2=\frac{38}{3} \approx 12.67$$ - If $$X_2=0$$, then $$2X_1=38 \Rightarrow X_1=19$$ 4. **Find intersection points of the constraints to identify vertices:** - Intersection of constraints 1 and 2: Solve system: $$40X_1 + 20X_2 = 600$$ $$4X_1 + 10X_2 = 100$$ Multiply second equation by 10: $$40X_1 + 100X_2 = 1000$$ Subtract first equation: $$40X_1 + 100X_2 - (40X_1 + 20X_2) = 1000 - 600$$ $$80X_2 = 400 \Rightarrow X_2 = 5$$ Substitute back: $$4X_1 + 10(5) = 100 \Rightarrow 4X_1 + 50 = 100 \Rightarrow 4X_1 = 50 \Rightarrow X_1 = 12.5$$ - Intersection of constraints 2 and 3: Solve system: $$4X_1 + 10X_2 = 100$$ $$2X_1 + 3X_2 = 38$$ Multiply second equation by 2: $$4X_1 + 6X_2 = 76$$ Subtract from first: $$(4X_1 + 10X_2) - (4X_1 + 6X_2) = 100 - 76$$ $$4X_2 = 24 \Rightarrow X_2 = 6$$ Substitute back: $$4X_1 + 10(6) = 100 \Rightarrow 4X_1 + 60 = 100 \Rightarrow 4X_1 = 40 \Rightarrow X_1 = 10$$ - Intersection of constraints 1 and 3: Solve system: $$40X_1 + 20X_2 = 600$$ $$2X_1 + 3X_2 = 38$$ Multiply second equation by 10: $$20X_1 + 30X_2 = 380$$ Multiply first equation by 1: $$40X_1 + 20X_2 = 600$$ Multiply second equation by 2: $$40X_1 + 60X_2 = 760$$ Subtract first equation: $$(40X_1 + 60X_2) - (40X_1 + 20X_2) = 760 - 600$$ $$40X_2 = 160 \Rightarrow X_2 = 4$$ Substitute back: $$2X_1 + 3(4) = 38 \Rightarrow 2X_1 + 12 = 38 \Rightarrow 2X_1 = 26 \Rightarrow X_1 = 13$$ 5. **List all vertices including intercepts and origin:** - $$A = (0,0)$$ - $$B = (0,10)$$ from constraint 2 intercept - $$C = (0,12.67)$$ from constraint 3 intercept - $$D = (15,0)$$ from constraint 1 intercept - $$E = (19,0)$$ from constraint 3 intercept - $$F = (12.5,5)$$ intersection of constraints 1 and 2 - $$G = (10,6)$$ intersection of constraints 2 and 3 - $$H = (13,4)$$ intersection of constraints 1 and 3 6. **Check which vertices satisfy all constraints (feasible region):** - Check each vertex against all constraints. - $$A(0,0)$$: satisfies all. - $$B(0,10)$$: - Constraint 1: $$40(0)+20(10)=200 \leq 600$$ OK - Constraint 3: $$2(0)+3(10)=30 \leq 38$$ OK - $$C(0,12.67)$$: - Constraint 1: $$20*12.67=253.4 \leq 600$$ OK - Constraint 2: $$4*0+10*12.67=126.7 \leq 100$$ NO (violates) - $$D(15,0)$$: - Constraint 2: $$4*15+10*0=60 \leq 100$$ OK - Constraint 3: $$2*15+3*0=30 \leq 38$$ OK - $$E(19,0)$$: - Constraint 1: $$40*19=760 \leq 600$$ NO (violates) - $$F(12.5,5)$$: - Check all constraints: - Constraint 3: $$2*12.5 + 3*5 = 25 + 15 = 40 \leq 38$$ NO (violates) - $$G(10,6)$$: - Constraint 1: $$40*10 + 20*6 = 400 + 120 = 520 \leq 600$$ OK - Constraint 3: $$2*10 + 3*6 = 20 + 18 = 38 \leq 38$$ OK - $$H(13,4)$$: - Constraint 2: $$4*13 + 10*4 = 52 + 40 = 92 \leq 100$$ OK - Constraint 3: $$2*13 + 3*4 = 26 + 12 = 38 \leq 38$$ OK 7. **Feasible vertices are:** $$A(0,0), B(0,10), D(15,0), G(10,6), H(13,4)$$ 8. **Evaluate objective function $$Z=40X_1 + 32X_2$$ at each feasible vertex:** - $$A: Z=40*0 + 32*0 = 0$$ - $$B: Z=40*0 + 32*10 = 320$$ - $$D: Z=40*15 + 32*0 = 600$$ - $$G: Z=40*10 + 32*6 = 400 + 192 = 592$$ - $$H: Z=40*13 + 32*4 = 520 + 128 = 648$$ 9. **Conclusion:** The maximum value of $$Z$$ is $$648$$ at $$X_1=13, X_2=4$$. **Final answer:** $$\boxed{Z_{max} = 648 \text{ at } (X_1, X_2) = (13,4)}$$