Discussion
If the right hand side of the second constraint is changed from 8 to 20, then in the optimal solution of the primal problem, the basic variables will be
Answer: D
The new optimal basis is \(x_{2}\) and \(x_{4}\).
With \(b = (4, 20)\), the current basis gives \(B^{-1}b\):
\[ \begin{aligned} x_{3} &= 4 - \tfrac{1}{4}(20) = -1 \\ x_{2} &= \tfrac{1}{4}(20) = 5 \end{aligned} \]
\(x_{3}\) is negative, so apply the dual simplex method with \(x_{3}\) leaving. In its row only the \(x_{4}\) entry, \(-1/4\), is negative, so \(x_{4}\) enters. Pivoting gives \(x_{4} = 4\) and \(x_{2} = 5 - \frac{1}{4}(4) = 4\), and the new \(z_{j} - c_{j}\) row is \((24, 0, 24, 0)\), all non-negative.
Check: \(x_{2} = 4\) satisfies the first constraint, and \(4(4) + 4 = 20\) the second, with \(z = 48 - 24 = 24\). Options keeping \(x_{3}\) are wrong because it became negative, and \(x_{1}\) never enters: its reduced cost becomes 24.
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