= Solution
In the $(u,v)$ plane the feasible set lies in the first quadrant above the three lines $v=3/2$, $v=5/2-2u$ and $v=2-(2/3)u$. Its lower boundary has vertices
$$
A=(3/4,3/2),\qquad B=(3/8,7/4),\qquad C=(0,5/2),
$$
and the region is unbounded upward and to the right. The objective at these vertices is respectively $21$, $43/2$ and $55/2$, so $A$ is optimal for the original supply.
\Image[/past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011-as2-dual-feasible.png]
{title=Dual feasible region and its three finite boundary vertices}
Let the available amount of R be $r\geq0$. Comparing the three dual <vertex> objectives gives the optimal value
$$
\boxed{V(r)=\begin{cases}
3r/4+33/2,&0\leq r\leq22/3,\\
3r/8+77/4,&22/3\leq r\leq22,\\
55/2,&r\geq22.
\end{cases}}
$$
Thus the marginal shadow price of R is $3/4$ below $22/3$, then $3/8$ between $22/3$ and $22$, and zero beyond $22$. At the two breakpoints, optimal dual prices range between the adjacent one-sided prices: $[3/8,3/4]$ and $[0,3/8]$ respectively. At zero supply the right marginal value is $3/4$; dual multipliers on the zero resource can be nonunique. These slope changes describe why the first extra $4/3$ units from the original supply six have the original price, but subsequent units do not.
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