Solution
= Solution
{c}
The basis and reduced costs stay unchanged when only the right-hand side changes. Its new basic values are
$$
\begin{pmatrix}x_3\\x_1\end{pmatrix}
=B^{-1}\begin{pmatrix}6\\4+\alpha\end{pmatrix}
=\begin{pmatrix}(2-\alpha)/3\\(14+2\alpha)/3\end{pmatrix}.
$$
Thus for small $|\alpha|$,
$$
\boxed{x_1=(14+2\alpha)/3,\quad x_3=(2-\alpha)/3,\quad x_2=x_4=x_5=0,\quad
z_{\max}=(26+5\alpha)/3.}
$$