The marginal value theorem holds true under three fairly mild conditions:

1. The fixed cost $T$ is larger than zero.
2. The reward function $E(t)$ increases with $t$.
3. The slope $dE(t)/dt$ of the reward function decreases with $t$ (i.e. $E(t)$ is a diminishing returns function).

We wish to prove that the instantaneous reward rate $r(t)$ equals the average reward rate $R(t)$ when $R(t)$ is maximal. To achieve this, we need to find the value of $r(t)$ when $R(t)$ is maximal. To find the maximal average reward rate, we make use of the fact that its slope is zero at a maximum. The average reward rate is defined as $$\begin{array}{rcl}R(t)& =& \frac{E(t)}{T+t},\end{array}$$ and its derivative is $$\begin{array}{rcl}\frac{dR}{dt}& =& \frac{1}{T+t}\frac{dE}{dt}+E(t)\frac{d(T+t{)}^{-1}}{dt},\end{array}$$ where (by definition) $$\begin{array}{rcl}\frac{dE}{dt}& =& r(t),\end{array}$$ is the instantaneous reward rate, and where $$\begin{array}{rcl}\frac{d((T+t{)}^{-1})}{dt}& =& \frac{-1}{(T+t{)}^2}.\end{array}$$ Substituting Equations $\text{???}$ and $\text{???}$ into Equation $\text{???}$, $$\begin{array}{rcl}\frac{dR}{dt}& =& r(t)\frac{1}{T+t}-\frac{E(t)}{(T+t{)}^2},\end{array}$$ where $$\begin{array}{rcl}\frac{E(t)}{T+t}& =& R(t),\end{array}$$ is the average reward rate, so that Equation $\text{???}$ becomes $$\begin{array}{rcl}\frac{dR}{dt}& =& \frac{r(t)}{T+t}-\frac{R(t)}{T+t}.\end{array}$$ At a maximum, this is equal to zero, $$\begin{array}{rcl}\frac{r(t)}{T+t}-\frac{R(t)}{T+t}& =& 0.\end{array}$$ Finally, multiplying both sides by $(T+t)$, and re-arranging yields $$\begin{array}{rcl}r(t)& =& R(t).\end{array}$$ This proves that the average reward rate is maximal when the instantaneous reward rate equals the average reward rate.

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