1. **State the problem:** Maximize the objective function $$P = 2x_1 + 5x_2$$ subject to the constraint $$-2x_1 + x_2 \leq 8$$ and $$x_1, x_2 \geq 0$$. 2. **Set up the problem for the simplex method:** Rewrite the constraint as $$-2x_1 + x_2 + s = 8$$ where $$s \geq 0$$ is the slack variable. 3. **Initial simplex tableau:** \[ \begin{array}{c|ccc|c} & x_1 & x_2 & s & RHS \\ \hline s & -2 & 1 & 1 & 8 \\ P & -2 & -5 & 0 & 0 \\ \end{array} \] 4. **Simplex method steps:** - Identify entering variable: most negative coefficient in objective row is $$-5$$ for $$x_2$$. - Identify leaving variable: ratio test $$\frac{8}{1} = 8$$ for slack variable $$s$$. - Pivot on $$x_2$$ in row 1. 5. **Pivot operation:** Make pivot element 1 (already 1), then eliminate $$x_2$$ from other rows. New tableau: \[ \begin{array}{c|ccc|c} & x_1 & x_2 & s & RHS \\ \hline x_2 & -2 & 1 & 1 & 8 \\ P & -2 & -5 & 0 & 0 \\ \end{array} \] Pivot row is row 1, so express $$x_2$$: $$x_2 = 8 + 2x_1 - s$$ Substitute into objective row: $$P = 2x_1 + 5x_2 = 2x_1 + 5(8 + 2x_1 - s) = 2x_1 + 40 + 10x_1 - 5s = 12x_1 + 40 - 5s$$ Rewrite objective row: $$P - 12x_1 + 5s = 40$$ 6. **Update tableau:** \[ \begin{array}{c|ccc|c} & x_1 & x_2 & s & RHS \\ \hline x_2 & -2 & 1 & 1 & 8 \\ P & 12 & 0 & 5 & 40 \\ \end{array} \] 7. **Check for optimality:** All coefficients in objective row for non-basic variables are non-negative (12 for $$x_1$$ and 5 for $$s$$), so not optimal yet. 8. **Next pivot:** Entering variable: $$x_1$$ (coefficient 12). Ratio test for leaving variable: From row 1: $$\frac{8}{-2}$$ is negative, so no positive ratio. Since no positive ratio, the problem is unbounded in $$x_1$$ direction. 9. **Graphical method:** Plot the constraint $$-2x_1 + x_2 \leq 8$$ and non-negativity constraints. Rewrite constraint: $$x_2 \leq 8 + 2x_1$$ The feasible region is below the line $$x_2 = 8 + 2x_1$$ in the first quadrant. Since $$x_1, x_2 \geq 0$$, the feasible region is unbounded extending to the right and upwards. 10. **Objective function lines:** Lines of constant $$P$$ are $$2x_1 + 5x_2 = c$$. Increasing $$P$$ moves the line outward in the direction of vector $$(2,5)$$. Because the feasible region is unbounded in the direction of increasing $$x_1$$ and $$x_2$$, and the objective function increases in that direction, the maximum is unbounded. **Final conclusion:** The linear program is unbounded; there is no finite maximum value for $$P$$.