= Solution
Apply the <Charnes-Cooper transformation> to any feasible point $x$ of $Q$:
$$
y=\frac{x}{d^Tx},
\qquad
t=\frac1{d^Tx}.
$$
Then
$$
y\geq0,\quad t>0,\quad Ay=bt,\quad d^Ty=1,
$$
so $(y,t)$ is feasible for $R$, and
$$
c^Ty=\frac{c^Tx}{d^Tx}.
$$
In particular, the given solution $x^*$ of $Q$ produces a feasible point of $R$ with the same objective value, so
$$
\max R\geq\max Q.
$$
Moreover, because every $d_i>0$, the equation $d^Ty=1$ implies
$$
0\leq y_i\leq\frac1{d_i}.
$$
The linear objective $c^Ty$ is therefore bounded on the feasible set. The feasible $y$ lie in the compact simplex $\{y\geq0:d^Ty=1\}$. Their subset arising in $R$ is closed: if $b\ne0$, any nonzero component of $b$ determines $t$ continuously from $Ay=bt$, while if $b=0$ the condition is simply $Ay=0$. Hence the feasible $y$ form a <compact set>, on which the continuous objective $c^Ty$ attains a finite maximum. Thus $R$ has a finite maximum at least as large as that of $Q$.
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