= Solution
The dual prices $u,v\geq0$ solve
$$
\boxed{\min 6u+11v,\qquad4u+2v\geq5,\quad2u+3v\geq6,\quad2v\geq3.}
$$
The optimal tableau gives $(u,v)=c_B^TB^{-1}=(6,3)B^{-1}=(3/4,3/2)$. This <vector> is dual feasible and has objective $21$, equal to the primal value, proving optimality.
Thus an extra unit of R is worth $3/4$ while the current <basis> stays feasible. Increasing its supply by $\delta$ changes the basic values to
$$
B^{-1}\begin{pmatrix}6+\delta\\11\end{pmatrix}
=\begin{pmatrix}3+\delta/2\\1-3\delta/4\end{pmatrix}.
$$
For increases, feasibility is retained exactly up to $\delta=4/3$. Therefore the firm can buy \b[up to $4/3$ extra units at a marginal value of $3/4$ per unit]. The reduced profits do not change during this right-hand-side variation.
For the proposed fourth good, the reduced profit is $8-(4u+4v)=8-9=-1$. Its profit is below the resource opportunity cost, so \b[it is not worth producing at the current optimum]. The old dual solution also satisfies the new good's inequality, certifying that introducing it cannot improve the optimum.
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