1. **Problem Statement:** We have the system of inequalities: $$\begin{cases} x_1 - x_2 + x_3 - 2x_4 \leq 3 \\ x_1 - 2x_2 + 4x_3 + 4x_4 \leq 4 \\ x_1, x_2, x_3, x_4 \geq 0 \end{cases}$$ We want to answer three questions about basic variables and basic solutions. 2. **Background:** - A basic solution in linear programming corresponds to selecting a subset of variables (basic variables) equal in number to the number of constraints, solving for them, and setting the rest (non-basic variables) to zero. - Here, we have 2 inequality constraints (converted to equalities with slack variables) and 4 variables. - The number of basic variables equals the number of constraints, so typically 2. --- **(a) How many basic variables does any basic solution have?** 3. Since there are 2 constraints, the number of basic variables is 2. **Answer:** at most 2 --- **(b) If $x_1, x_2$ are basic variables, find the corresponding basic solution.** 4. Set non-basic variables $x_3 = 0$, $x_4 = 0$. 5. The system reduces to: $$\begin{cases} x_1 - x_2 \leq 3 \\ x_1 - 2x_2 \leq 4 \end{cases}$$ 6. Treat inequalities as equalities for basic solution: $$\begin{cases} x_1 - x_2 = 3 \\ x_1 - 2x_2 = 4 \end{cases}$$ 7. Subtract second from first: $$ (x_1 - x_2) - (x_1 - 2x_2) = 3 - 4 \Rightarrow -x_2 + 2x_2 = -1 \Rightarrow x_2 = -1 $$ 8. Since $x_2 = -1$ is negative, this violates $x_2 \geq 0$. 9. So no feasible basic solution with $x_1, x_2$ basic and $x_3 = x_4 = 0$. 10. Check given options for $x_1, x_2$ basic: - $(1, 5/2, -1/2, 0)$ has $x_3 = -1/2 < 0$, invalid. - $(0,1,0,1)$ has $x_1=0$, $x_2=1$, but $x_4=1$ nonzero, so $x_4$ is not zero. - $(1,0,1,0)$ has $x_2=0$, $x_3=1$ nonzero. - $(1,2,0,0)$ has $x_1=1$, $x_2=2$, $x_3=0$, $x_4=0$. - $(2,1,0,0)$ has $x_1=2$, $x_2=1$, $x_3=0$, $x_4=0$. 11. Check if $(1,2,0,0)$ satisfies constraints: - $1 - 2 + 0 - 0 = -1 \leq 3$ true. - $1 - 4 + 0 + 0 = -3 \leq 4$ true. 12. Check if $(2,1,0,0)$ satisfies constraints: - $2 - 1 + 0 - 0 = 1 \leq 3$ true. - $2 - 2 + 0 + 0 = 0 \leq 4$ true. 13. Both satisfy constraints and non-negativity. 14. Since $x_3$ and $x_4$ are zero, and $x_1, x_2$ are basic variables, the basic solution is $(1,2,0,0)$ or $(2,1,0,0)$. 15. Among options, $(1,2,0,0)$ is listed. **Answer:** $(x_1,x_2,x_3,x_4)^T = (1, 2, 0, 0)$ --- **(c) If $x_2, x_4$ are basic variables, find the corresponding basic solution.** 16. Set non-basic variables $x_1 = 0$, $x_3 = 0$. 17. The system reduces to: $$\begin{cases} 0 - x_2 + 0 - 2x_4 \leq 3 \\ 0 - 2x_2 + 0 + 4x_4 \leq 4 \end{cases}$$ 18. Treat as equalities: $$\begin{cases} -x_2 - 2x_4 = 3 \\ -2x_2 + 4x_4 = 4 \end{cases}$$ 19. Solve system: From first: $-x_2 - 2x_4 = 3 \Rightarrow x_2 = -2x_4 - 3$ Substitute into second: $$-2(-2x_4 - 3) + 4x_4 = 4 \Rightarrow 4x_4 + 6 + 4x_4 = 4 \Rightarrow 8x_4 + 6 = 4 \Rightarrow 8x_4 = -2 \Rightarrow x_4 = -\frac{1}{4}$$ 20. Since $x_4 = -\frac{1}{4} < 0$, violates $x_4 \geq 0$. 21. So no feasible basic solution with $x_2, x_4$ basic and $x_1 = x_3 = 0$. 22. Check options: - $(0,1,0,1)$: $x_2=1$, $x_4=1$, $x_1=0$, $x_3=0$. Check constraints: $0 - 1 + 0 - 2(1) = -3 \leq 3$ true. $0 - 2(1) + 0 + 4(1) = 2 \leq 4$ true. - $(0,1,0,2)$: check first constraint: $0 - 1 + 0 - 4 = -5 \leq 3$ true. Second: $0 - 2 + 0 + 8 = 6 \leq 4$ false. - $(0,2,0,1)$: first: $0 - 2 + 0 - 2 = -4 \leq 3$ true. Second: $0 - 4 + 0 + 4 = 0 \leq 4$ true. 23. But $x_2=2$, $x_4=1$ with $x_1=x_3=0$ is feasible. 24. So $(0,2,0,1)$ is a feasible basic solution. 25. However, from step 19, the system solution for equalities is not feasible. 26. Since basic solution requires equalities, and no solution satisfies equalities with $x_2,x_4$ basic, the basic solution does not exist. **Answer:** does not exist --- **Summary:** - (a) at most 2 - (b) $(1,2,0,0)$ - (c) does not exist