min x1+1/3(x2+1)^2,

subject to:

x1 + 2x2 = 4

x1 - x2 >= 1

where x1 and x2 are real-valued decision variables.

This is a linear programming problem. We can use the simplex algorithm to solve it.

First, we can convert the problem to standard form by introducing slack variables:

x1 + 2x2 + s1 = 4

x1 - x2 - s2 = -1

The problem now becomes:

minimize z = x1 + 1/3(x2+1)^2

subject to:

x1 + 2x2 + s1 = 4

x1 - x2 - s2 = -1

where x1, x2, s1, s2 are non-negative decision variables.

Next, we can introduce the artificial variables y1 and y2, and convert the problem to auxiliary form:

minimize z = x1 + 1/3(x2+1)^2

subject to:

x1 + 2x2 + s1 + y1 = 4

x1 - x2 - s2 + y2 = -1

where x1, x2, s1, s2, y1, y2 are non-negative decision variables.

We can then apply the simplex algorithm to solve the problem. The initial tableau is:

z | x1 | x2 | s1 | s2 | y1 | y2 | RHS
-------------------------------------
1 | 1 | 1/3| 0 | 0 | 0 | 0 | 0
-------------------------------------
0 | 1 | 2 | 1 | 0 | 1 | 0 | 4
-------------------------------------
0 | 1 | -1 | 0 | -1 | 0 | 1 | -1

The pivot column is x2, and the pivot row is the second row. We perform the pivot operation to make x2 the basic variable in the second equation:

z | x1 | x2 | s1 | s2 | y1 | y2 | RHS
-------------------------------------
1 | 1 | 1/3| 0 | 0 | 0 | 0 | 0
-------------------------------------
0 | 1 | 2 | 1 | 0 | 1 | 0 | 4
-------------------------------------
0 | 1 | 1 | 0 | 1 | 0 | -1 | 1

We then perform the pivot operation to make x1 the basic variable in the first equation:

z | x1 | x2 | s1 | s2 | y1 | y2 | RHS
-------------------------------------
1 | 0 | 5/3| 0 | 0 | -1 | 0 | -4/3
-------------------------------------
0 | 1 | 0 | 1/3| -2 | 1/3| -2/3| 10/3
-------------------------------------
0 | 0 | 1 | 0 | 1 | 0 | -1 | 1

The optimal solution is x1 = 4/3, x2 = 1, z = 5/3. The constraint x1 - x2 >= 1 is satisfied with equality, which means that the optimal solution lies on the boundary of the feasible region.