1.
a) x1 = 10, x2 = 40, s3 = 30, z = 420 b) Yes; all cj – zj row values are zero or negative. c) x3 = 0; s2 = 0 d) Maximize Z = 10x1 + 2x2 + 6x3 e) 3 f) Since there are three decision variables, a three-dimensional graph is required.
2.
a) minimization; because zj – cj is being calculated on the bottom row and not cj – zj b) x1 = 20, x3 = 10, s1 = 10, z = 240 c) Yes; all zj – cj values are negative or zero. d) Minimize Z = 6x1 + 20x2 + 12x3 e) 3 constraints f) Yes, one; because there are 3 constraints but only 2 surplus variables. Since both ≤ and ≥ constraints have slack or surplus variables, and an equality does not, then one of the three constraints was an =. g) x2 = 0
3.
a) maximization; because of “cj-zj” b) x1 = 12, x2 = 0, x3 = 0, x4 = 15 c) s1 = 20, s2 = 0, s3 = 0, s4 = 45 d) If x2 is selected as the entering basic variable, Z will increase by 20 for every unit of x2 entered into the solution. e) This solution is not optimal because there are positive values in the cj - zj row.