= Solution
Let $(y,t)$ be feasible for $R$. Since $d_i>0$ and $d^Ty=1$, the vector $y$ is nonzero. Every entry of $A$ is strictly positive and $y\geq0$, so every component of $Ay$ is strictly positive. The relation
$$
Ay=bt
$$
therefore forces $t>0$.
Set $x=y/t$. Then
$$
x\geq0,\qquad Ax=b,\qquad d^Tx=\frac1t>0,
$$
so $x$ is feasible for $Q$, and
$$
\frac{c^Tx}{d^Tx}
=\frac{c^Ty/t}{1/t}
=c^Ty.
$$
Thus every feasible value of $R$ is a feasible value of the <linear-fractional programming>[linear-fractional program] $Q$. Together with part (i), this proves
$$
\boxed{\max R=\max Q}.
$$
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