1. **Stating the problem:** Maximize the objective function $$z(x) = 8x_1 + 5x_2$$ subject to the constraints: $$\begin{cases} 2x_1 + x_2 \leq 10, \\ x_1 + x_2 \leq 12, \\ 4x_1 + x_2 \leq 18, \\ x_1 + 4x_2 \leq 10, \\ x_1 \geq 0, x_2 \geq 0 \end{cases}$$ 2. **Formula and rules:** This is a linear programming problem (PPL) where we want to maximize a linear function subject to linear inequalities. - The feasible region is the intersection of all inequalities. - The maximum of a linear function over a polygonal region occurs at a vertex (corner point) of the feasible region. 3. **Find vertices of the feasible region:** We find intersection points of the constraint lines and check which satisfy all inequalities. - Intersection of $2x_1 + x_2 = 10$ and $x_1 + x_2 = 12$: $$\text{Subtract: } (2x_1 + x_2) - (x_1 + x_2) = 10 - 12 \Rightarrow x_1 = -2$$ Not feasible since $x_1 \geq 0$. - Intersection of $2x_1 + x_2 = 10$ and $4x_1 + x_2 = 18$: $$\text{Subtract: } (4x_1 + x_2) - (2x_1 + x_2) = 18 - 10 \Rightarrow 2x_1 = 8 \Rightarrow x_1 = 4$$ Substitute back: $$2(4) + x_2 = 10 \Rightarrow 8 + x_2 = 10 \Rightarrow x_2 = 2$$ Check other constraints: $$4 + 2 = 6 \leq 12, \quad 4 + 4(2) = 4 + 8 = 12 \leq 10 \text{ (No, violates)}$$ So point $(4,2)$ is not feasible. - Intersection of $2x_1 + x_2 = 10$ and $x_1 + 4x_2 = 10$: From first: $x_2 = 10 - 2x_1$ Substitute into second: $$x_1 + 4(10 - 2x_1) = 10 \Rightarrow x_1 + 40 - 8x_1 = 10 \Rightarrow -7x_1 = -30 \Rightarrow x_1 = \frac{30}{7} \approx 4.29$$ Then: $$x_2 = 10 - 2 \times \frac{30}{7} = 10 - \frac{60}{7} = \frac{70 - 60}{7} = \frac{10}{7} \approx 1.43$$ Check other constraints: $$x_1 + x_2 = \frac{30}{7} + \frac{10}{7} = \frac{40}{7} \approx 5.71 \leq 12$$ $$4x_1 + x_2 = 4 \times \frac{30}{7} + \frac{10}{7} = \frac{120 + 10}{7} = \frac{130}{7} \approx 18.57 \leq 18 \text{ (No, violates)}$$ So not feasible. - Intersection of $x_1 + x_2 = 12$ and $4x_1 + x_2 = 18$: Subtract: $$4x_1 + x_2 - (x_1 + x_2) = 18 - 12 \Rightarrow 3x_1 = 6 \Rightarrow x_1 = 2$$ Then: $$x_2 = 12 - 2 = 10$$ Check $x_1 + 4x_2 = 2 + 40 = 42 \leq 10$? No, violates. - Intersection of $x_1 + x_2 = 12$ and $x_1 + 4x_2 = 10$: From first: $x_1 = 12 - x_2$ Substitute into second: $$12 - x_2 + 4x_2 = 10 \Rightarrow 12 + 3x_2 = 10 \Rightarrow 3x_2 = -2 \Rightarrow x_2 = -\frac{2}{3}$$ Not feasible since $x_2 \geq 0$. - Intersection of $4x_1 + x_2 = 18$ and $x_1 + 4x_2 = 10$: From second: $x_1 = 10 - 4x_2$ Substitute into first: $$4(10 - 4x_2) + x_2 = 18 \Rightarrow 40 - 16x_2 + x_2 = 18 \Rightarrow 40 - 15x_2 = 18 \Rightarrow -15x_2 = -22 \Rightarrow x_2 = \frac{22}{15} \approx 1.47$$ Then: $$x_1 = 10 - 4 \times \frac{22}{15} = 10 - \frac{88}{15} = \frac{150 - 88}{15} = \frac{62}{15} \approx 4.13$$ Check other constraints: $$2x_1 + x_2 = 2 \times \frac{62}{15} + \frac{22}{15} = \frac{124 + 22}{15} = \frac{146}{15} \approx 9.73 \leq 10$$ $$x_1 + x_2 = \frac{62}{15} + \frac{22}{15} = \frac{84}{15} = 5.6 \leq 12$$ All constraints satisfied. - Check intercepts: - At $x_1=0$: $$2(0) + x_2 \leq 10 \Rightarrow x_2 \leq 10$$ $$0 + x_2 \leq 12 \Rightarrow x_2 \leq 12$$ $$4(0) + x_2 \leq 18 \Rightarrow x_2 \leq 18$$ $$0 + 4x_2 \leq 10 \Rightarrow x_2 \leq \frac{10}{4} = 2.5$$ So max $x_2$ at $x_1=0$ is $2.5$. - At $x_2=0$: $$2x_1 + 0 \leq 10 \Rightarrow x_1 \leq 5$$ $$x_1 + 0 \leq 12 \Rightarrow x_1 \leq 12$$ $$4x_1 + 0 \leq 18 \Rightarrow x_1 \leq 4.5$$ $$x_1 + 0 \leq 10 \Rightarrow x_1 \leq 10$$ So max $x_1$ at $x_2=0$ is $4.5$. 4. **Vertices of feasible region:** - $(0,0)$ - $(0,2.5)$ - $(4.13,1.47)$ (intersection of $4x_1 + x_2 = 18$ and $x_1 + 4x_2 = 10$) - $(4.5,0)$ 5. **Evaluate objective function at vertices:** - At $(0,0)$: $z = 8(0) + 5(0) = 0$ - At $(0,2.5)$: $z = 8(0) + 5(2.5) = 12.5$ - At $(4.13,1.47)$: $$z = 8 \times 4.13 + 5 \times 1.47 = 33.04 + 7.35 = 40.39$$ - At $(4.5,0)$: $z = 8(4.5) + 5(0) = 36$ 6. **Conclusion:** The maximum value of $z$ is approximately $$\boxed{40.39}$$ at the point $$\left(\frac{62}{15}, \frac{22}{15}\right) \approx (4.13, 1.47).$$