Solution. \(B^{-1}e_1 = (1, \frac 23, -\frac 13)\), so the conditions are \(3 + \delta \geq 0\), \(8 + \frac 23\delta \geq 0\), \(10 - \frac 13\delta \geq 0\): \(-3 \leq \delta \leq 30\), that is \(23 \leq b_1 \leq 56\). At \(b_1 = 30\), \(\delta = 4\): \(s_3 = 7\), \(x_1 = 8 + \frac 83 = \frac {32}3\), \(x_2 = 10 - \frac 43 = \frac {26}3\), and \(Z = 230 + 4 \cdot \frac 53 = 236.7\).

Solution. With \(c_B = (0, c_1, 15)\) for \((s_3, x_1, x_2)\), the multipliers are \(y = c_B^{\mathsf T}B^{-1} = (\frac 23c_1 - 5, -\frac 16c_1 + 5, 0)\), and the reduced profits of \(s_1\) and \(s_2\) are \(-y_1\) and \(-y_2\). Optimality requires \(y_1 \geq 0\) and \(y_2 \geq 0\): \(c_1 \geq 7.5\) and \(c_1 \leq 30\). At \(c_1 = 10\) these give \(y = (\frac 53, \frac {10}3, 0)\).

Solution. It is worth making if its profit exceeds \(y^{\mathsf T}a = 3\cdot \frac 53 + 1\cdot \frac {10}3 + 0 = \frac {25}3 = 8.33\).

Solution. The allowed increase of \(b_1\) is \(30\) and the allowed decrease of \(b_2\) is \(30\). The fractions are \(\frac {10}{30} + \frac {15}{30} = \frac 56 \leq 1\), so the basis remains optimal. Check: \[ x_B = (3, 8, 10) + 10\bigl (1, \tfrac 23, -\tfrac 13\bigr ) - 15\bigl (-\tfrac 12, -\tfrac 16, \tfrac 13\bigr ) = (20.5, 17.17, 1.67) \geq 0 . \]

Solution. For a bounded feasible region with vertices \(v^1, \ldots , v^K\), the optimal value of the maximisation is \(\max _kc^{\mathsf T}v^k\), a maximum of linear functions of \(c\), which is convex. Economically, if prices can vary, being able to re-optimise is worth more the more they vary.

Solution. No. Its shadow price is \(0\) for all \(b_3 \geq 2\), so extra units add no profit: the firm would pay Rs. 10 for nothing. The third resource is not a bottleneck; the first two are, worth \(\frac 53\) and \(\frac {10}3\) per unit within their ranges.