Solve using artificial variables. \[ \begin{array}{lc} \text { Maximize } & z=3 x_{1}+2 x_{2} \\ \text { subject to: } & x_{1}+x_{2}=70 \\ & 4 x_{1}+2 x_{2} \geq 120 \\ & 5 x_{1}+2 x_{2} \leq 250 \\ & x_{1} \geq 0, x_{2} \geq 0 \end{array} \] The maximum is $z=$ when $\mathrm{x}_{1}=$ and $x_{2}=$ (Simplify your answers.)

Step 1 ：Convert the inequalities into equalities by introducing slack, surplus and artificial variables. The inequality \(4x1 + 2x2 \geq 120\) can be rewritten as \(4x1 + 2x2 - s1 = 120\), where \(s1\) is a surplus variable. The inequality \(5x1 + 2x2 \leq 250\) can be rewritten as \(5x1 + 2x2 + s2 = 250\), where \(s2\) is a slack variable. Since we have a surplus variable in the first constraint, we need to introduce an artificial variable \(a1\). So, the first constraint becomes \(4x1 + 2x2 - s1 + a1 = 120\).

Step 2 ：Form the initial simplex tableau. The initial simplex tableau is: \[\begin{array}{cccccc} & x1 & x2 & s1 & a1 & RHS \\ a1 & 4 & 2 & -1 & 1 & 120 \\ s2 & 5 & 2 & 0 & 1 & 250 \\ z & -3 & -2 & 0 & 0 & 0 \end{array}\]

Step 3 ：Perform the simplex method. The pivot column is the one with the most negative coefficient in the bottom row, which is \(x1\). The pivot row is the one with the smallest non-negative ratio of the right-hand side to the pivot column, which is the first row. So, the pivot element is 4. After performing the pivot operation, we get the following tableau: \[\begin{array}{cccccc} & x1 & x2 & s1 & a1 & RHS \\ x1 & 1 & 0.5 & -0.25 & 0.25 & 30 \\ s2 & 0 & 1 & 0.25 & -1.25 & 100 \\ z & 0 & 1 & 0.75 & 0.75 & 90 \end{array}\]

Step 4 ：Check for optimality. Since all the coefficients in the bottom row are non-negative, we have reached the optimal solution. The maximum value of \(z\) is 90 when \(x1 = 30\) and \(x2 = 40\). \(\boxed{z_{max} = 90}\)