Solve the following problem using x3 and x4 as starting basic feasible variables. As in Problem 3-39, do not use any artificial variables.

Minimize z = 3x1 + 2x2 + 3x3 subject to x1 + 4x2 + x3 Ú 14 2x1 + x2 + x4 Ú 20 x1, x2, x3, x4 Ú 0

*3-41. Consider the problem Maximize z = x1 + 5x2 + 3x3 subject to x1 + 2x2 + x3 = 6 2x1 - x2 = 8 x1, x2, x3 Ú 0 The variable x3 plays the role of a slack. Thus, no artificial variable is needed in the first constraint. In the second constraint, an artificial variable, R, is needed. Solve the problem using x3 and R as the starting variables.