1. **State the problem:** We want to maximize the objective function $$Z = 3X + 5Y$$ subject to the constraints: $$3X + 5Y \leq 100$$ $$8X + 6Y \leq 80$$ with $$X \geq 0$$ and $$Y \geq 0$$. 2. **Identify the feasible region:** The feasible region is defined by the inequalities above. We will find the corner points by solving the system of equations formed by the constraints. 3. **Find intercepts for each constraint:** - For $$3X + 5Y = 100$$: - When $$X=0$$, $$5Y=100 \Rightarrow Y=20$$. - When $$Y=0$$, $$3X=100 \Rightarrow X=\frac{100}{3} \approx 33.33$$. - For $$8X + 6Y = 80$$: - When $$X=0$$, $$6Y=80 \Rightarrow Y=\frac{80}{6} = \frac{40}{3} \approx 13.33$$. - When $$Y=0$$, $$8X=80 \Rightarrow X=10$$. 4. **Find intersection point of the two constraints:** Solve the system: $$\begin{cases} 3X + 5Y = 100 \\ 8X + 6Y = 80 \end{cases}$$ Multiply the first equation by 6 and the second by 5 to eliminate $$Y$$: $$\begin{cases} 18X + 30Y = 600 \\ 40X + 30Y = 400 \end{cases}$$ Subtract the first from the second: $$40X + 30Y - (18X + 30Y) = 400 - 600$$ $$22X = -200$$ $$X = \frac{-200}{22} = -\frac{100}{11} \approx -9.09$$ Since $$X$$ cannot be negative, the intersection is outside the feasible region. 5. **Evaluate $$Z$$ at corner points of the feasible region:** - At $$X=0, Y=0$$: $$Z = 3(0) + 5(0) = 0$$ - At $$X=0, Y=13.33$$ (from second constraint): $$Z = 3(0) + 5(13.33) = 66.67$$ - At $$X=10, Y=0$$ (from second constraint): $$Z = 3(10) + 5(0) = 30$$ - At $$X=\frac{100}{3} \approx 33.33, Y=0$$ (from first constraint): $$Z = 3(33.33) + 5(0) = 100$$ - At $$X=0, Y=20$$ (from first constraint): $$Z = 3(0) + 5(20) = 100$$ 6. **Check feasibility of points:** - Point $$X=\frac{100}{3}, Y=0$$ satisfies: $$8(33.33) + 6(0) = 266.64 > 80$$ (not feasible) - Point $$X=0, Y=20$$ satisfies: $$8(0) + 6(20) = 120 > 80$$ (not feasible) - Point $$X=0, Y=13.33$$ satisfies: $$3(0) + 5(13.33) = 66.65 \leq 100$$ and $$8(0) + 6(13.33) = 80$$ (feasible) - Point $$X=10, Y=0$$ satisfies: $$3(10) + 5(0) = 30 \leq 100$$ and $$8(10) + 6(0) = 80$$ (feasible) - Point $$X=0, Y=0$$ is trivially feasible. 7. **Check points on the line segment between $$X=10, Y=0$$ and $$X=0, Y=13.33$$:** Since the intersection of constraints is not feasible, the maximum lies on the line segment between these two points. 8. **Calculate $$Z$$ at these points:** - At $$X=10, Y=0$$: $$Z=30$$ - At $$X=0, Y=13.33$$: $$Z=66.67$$ 9. **Conclusion:** The maximum value of $$Z$$ under the constraints is approximately $$66.67$$ at $$X=0, Y=\frac{40}{3}$$. **Final answer:** $$X=0, Y=\frac{40}{3} \approx 13.33, \quad Z_{max} = 66.67$$