1. **State the problem:** We want to maximize the objective function $$P = 35x_1 + x_2$$ subject to the constraints: $$2x_1 + x_2 \leq 8$$ $$x_1 + 5x_2 \leq 8$$ with $$x_1, x_2 \geq 0$$. 2. **Introduce slack variables:** To convert inequalities into equalities, add slack variables $$s_1$$ and $$s_2$$ for the first and second constraints respectively: $$2x_1 + x_2 + s_1 = 8$$ $$x_1 + 5x_2 + s_2 = 8$$ where $$s_1, s_2 \geq 0$$. 3. **Rewrite the objective function for the simplex method:** We write the objective function as: $$P - 35x_1 - x_2 = 0$$ 4. **Initial system with slack variables:** $$\begin{cases} 2x_1 + x_2 + s_1 = 8 \\ x_1 + 5x_2 + s_2 = 8 \\ P - 35x_1 - x_2 = 0 \end{cases}$$ 5. **Simplex tableau:** Set up the tableau with variables $$x_1, x_2, s_1, s_2, P$$: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline s_1 & 2 & 1 & 1 & 0 & 8 \\ s_2 & 1 & 5 & 0 & 1 & 8 \\ -P & -35 & -1 & 0 & 0 & 0 \end{array}$$ 6. **Identify entering and exiting variables:** - Entering variable: The variable with the most negative coefficient in the bottom row is $$x_1$$ (coefficient $$-35$$). - Compute ratios for pivot row: - For $$s_1$$: $$\frac{8}{2} = 4$$ - For $$s_2$$: $$\frac{8}{1} = 8$$ - Exiting variable: The smallest positive ratio is 4, so $$s_1$$ leaves the basis. - Pivot element is 2 in row 1, column $$x_1$$. 7. **Perform pivot operation:** Divide row 1 by pivot element 2: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline s_1 & \cancel{2}/2 & \frac{1}{2} & \frac{1}{2} & 0 & 4 \\ s_2 & 1 & 5 & 0 & 1 & 8 \\ -P & -35 & -1 & 0 & 0 & 0 \end{array} \Rightarrow \begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline x_1 & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 4 \\ s_2 & 1 & 5 & 0 & 1 & 8 \\ -P & -35 & -1 & 0 & 0 & 0 \end{array}$$ Subtract row 1 from row 2: $$\begin{aligned} s_2 & : (1 - 1)x_1 + (5 - \frac{1}{2})x_2 + (0 - \frac{1}{2})s_1 + (1 - 0)s_2 = 8 - 4 \\ & = 0x_1 + \frac{9}{2}x_2 - \frac{1}{2}s_1 + s_2 = 4 \end{aligned}$$ Update row 2: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline x_1 & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 4 \\ s_2 & 0 & \frac{9}{2} & -\frac{1}{2} & 1 & 4 \\ -P & -35 & -1 & 0 & 0 & 0 \end{array}$$ Add 35 times row 1 to row 3: $$\begin{aligned} -P & : (-35 + 35 \times 1)x_1 + (-1 + 35 \times \frac{1}{2})x_2 + (0 + 35 \times \frac{1}{2})s_1 + 0s_2 = 0 + 35 \times 4 \\ & = 0x_1 + ( -1 + 17.5 )x_2 + 17.5 s_1 + 0 = 140 \end{aligned}$$ Update row 3: $$\begin{array}{c|cccc|c} & x_1 & x_2 & s_1 & s_2 & RHS \\ \hline x_1 & 1 & \frac{1}{2} & \frac{1}{2} & 0 & 4 \\ s_2 & 0 & \frac{9}{2} & -\frac{1}{2} & 1 & 4 \\ -P & 0 & 16.5 & 17.5 & 0 & 140 \end{array}$$ 8. **Check for optimality:** All coefficients in the bottom row for variables $$x_1$$ and $$x_2$$ are now non-negative, so the current solution is optimal. 9. **Solution:** From the tableau: $$x_1 = 4, \quad x_2 = 0, \quad s_1 = 0, \quad s_2 = 4$$ Maximum value of $$P$$ is: $$P = 35(4) + 0 = 140$$ **Final answer:** $$x_1 = 4, x_2 = 0, P_{max} = 140$$