Proposition 2 — the constant rate λ^D = min{λ*, x/t} solves the deterministic problem (11); J^D(x,t) = t r(λ^D)

Let $(\Lambda,p,\lambda^*)$ be a regular demand function, $x\ge0$ a stock and $t>0$ a horizon. Write $\lambda^0=x/t$ for the run-out rate, $p^0=p(\lambda^0)$ and $r^0=p^0\lambda^0$, and set

Then:

1. the constant path $\lambda(s)=\lambda^D$, $0\le s\le t$, is feasible for the deterministic problem (11);
2. it is optimal: every feasible rate path $\lambda(\cdot)$ satisfies $\int_0^t r(\lambda(s))\,ds\le t\,r(\lambda^D)$;
3. the optimal revenue is

which equals $t\min\{r^*,r^0\}$ when $x\le\lambda^*t$, and $t\,r^*$ when $x\ge\lambda^*t$; 4. in terms of price, $p^D=p(\lambda^D)=\max\{p^*,p^0\}$ (when $\lambda^*>0$, $x>0$ and $\lambda^0\in\Lambda$, so that both prices are finite).

If items are plentiful ($x\ge\lambda^*t$) the firm prices at the revenue-maximizing level; if they are scarce it prices to sell exactly its stock. This solution supplies the fixed-price heuristic and the value against which Theorems 2 and 3 compare the stochastic problem.

Formalization Note The page prints $J^D(x,t)=t\min\{r^*,r^0\}$ (eq. (12)) without a case distinction. That is wrong when $x>\lambda^*t$: there $r^0=r(\lambda^0)\le r^*$, so $t\min\{r^*,r^0\}=t\,r^0$, while the optimum is $t\,r^*$. For exponential demand $ae^{-p}$ with $x=at$, $r^0=0$ but $J^D=a t/e$. The statement gives the corrected value $t\,r(\lambda^D)$, and the printed form in the case $x\le\lambda^*t$, where it is right. The page calls $\lambda^D$ "the" optimal solution; under concavity alone it need not be unique, so the statement asserts optimality, not uniqueness. When $\lambda^0\notin\Lambda$ the run-out quantities $p^0$, $r^0$ are undefined, and the statement uses them only where $\lambda^0\in\Lambda$. The OCR's "$p(\lambda_0)$" is a typesetting slip for $p(\lambda^0)$.