= Solution
The old <slack variable> $s_3$ is exactly the amount now sold, so rename its column $y$. No further slack is added: the PDF specifies an equality. The feasible equations are otherwise unchanged. If $z_0$ is the old profit, the new profit is $z=z_0+y$. Substituting this into the old objective equation gives
$$
z+s_1-\frac12y=\frac{27}{2}.
$$
This is the <simplex objective update for a priced slack variable>. The altered <simplex tableau> is
$$
\begin{array}{c|rrrrr|r}
\text{basis}&x_1&x_2&s_1&s_2&y&\text{RHS}\\\hline
x_1&1&0&-1&0&1/2&3/2\\
x_2&0&1&1&0&-1/4&9/4\\
s_2&0&0&1&1&-3/4&3/4\\\hline
z&0&0&1&0&-1/2&27/2
\end{array}.
$$
The coefficient $-1/2$ makes $y$ an improving entering variable. Its only positive constraint-row coefficient is $1/2$ in the $x_1$ row, whose ratio is $(3/2)/(1/2)=3$. Thus $x_1$ leaves. The <simplex method> pivot gives
$$
\begin{array}{c|rrrrr|r}
\text{basis}&x_1&x_2&s_1&s_2&y&\text{RHS}\\\hline
y&2&0&-2&0&1&3\\
x_2&1/2&1&1/2&0&0&3\\
s_2&3/2&0&-1/2&1&0&3\\\hline
z&1&0&0&0&0&15
\end{array}.
$$
The objective equation $z=15-x_1$ proves optimality. One optimal basic solution is
$$
\boxed{x_1=0,\qquad x_2=3,\qquad y=3,\qquad z_{\max}=15.}
$$
The zero reduced cost of $s_1$ permits alternative optima. In fact all optimal solutions are
$$
\boxed{x_1=0,\qquad 0\leq x_2\leq3,\qquad y=15-4x_2.}
$$
These satisfy both remaining inequalities, and the equality then makes their profit exactly fifteen.
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