Solution. The vertices are \(A(\frac 52, \frac 14)\), \(B(\frac 52, \frac 32)\) and \(C(0, \frac 32)\), with \(Z = \frac {73}4\), \(22\) and \(\frac 92\). The maximum is \(22\) at \(B\). Three constraint lines pass through \(B\): \(x_1 = \frac 52\), \(x_2 = \frac 32\) and \(x_1 + x_2 = 4\). In standard form, with four slack or surplus variables and \(m = 4\), the BFS at \(B\) has \(x_1, x_2\) and the surplus of the first constraint positive and the slacks of the other three zero, so one basic variable is zero: degenerate, and \(B\) is given by three different bases.

Solution. Take \(x_2 = -t\) with \(t \geq 3\) and \(x_1 = 4 + 2t\). Then \(-3x_1 + x_2 = -12 - 7t \leq 6\) and \(x_1 + 2x_2 = 4\), so the point is feasible, and the objective is \(-4 - 2t - 4t = -4 - 6t \to -\infty \). The region contains the ray \((4, 0) + t(2, -1)\), \(t \geq 3\), along which the objective decreases; Software 2.4.2 reports status \(3\).

Solution. If \(x\) and \(y\) are optimal with value \(z^*\), then \(\lambda x + (1 - \lambda )y\) is feasible, since the feasible region is convex, and its objective value is \(\lambda z^* + (1 - \lambda )z^* = z^*\) by linearity, so it is optimal.

Solution. There are \(\binom 42 = 6\) choices of basis. \(\{x_1, x_2\}\): \(x = (3, 1, 0, 0)\), feasible. \(\{x_1, x_3\}\): \(x_1 = 2\), \(x_3 = 2\), feasible. \(\{x_1, x_4\}\): \(x_1 = 4\), \(x_4 = -1\), infeasible. \(\{x_2, x_3\}\): \(x_2 = -2\), infeasible. \(\{x_2, x_4\}\): \(x_2 = 4\), \(x_4 = 3\), feasible. \(\{x_3, x_4\}\): \(x_3 = 4\), \(x_4 = 1\), feasible. None is degenerate, since every basic variable in the feasible ones is positive.

Solution. \(B\) is optimal while the slope of the iso-profit lines, \(-10/c_2\), lies between the slopes of the two constraints tight at \(B\): \(-2\) for \(2x_1 + x_2 = 26\) and \(-\frac 12\) for \(2x_1 + 4x_2 = 56\). So \(-2 \leq -10/c_2 \leq -\frac 12\), that is \(5 \leq c_2 \leq 20\). At \(c_2 = 20\) the edge \(BC\) is also optimal, and above it \(C\) takes over; at \(c_2 = 5\) the edge \(AB\) is optimal.

Solution. The vertices are \((1, 0)\) and \((0, 1)\), and the unbounded edges point in the directions \((1, 0)\) and \((0, 1)\). Then \((3, 2) = 1\cdot (1, 0) + 2\cdot (1, 0) + 2\cdot (0, 1)\), a convex combination of vertices (here just one, with \(\lambda = 1\)) plus \(\mu _1 = 2\), \(\mu _2 = 2\) times the rays. The representation is not unique: \(\frac 12(1, 0) + \frac 12(0, 1) + \frac 52(1, 0) + \frac 32(0, 1)\) also works.