I agree with some of the observations already done, but I disapprove the discrete approach so I'll follow a continous one.
If there is a moment when it is convenient to be in Area $2$, than it won't ever be convenient to go (back) to Area $1$: time spent in Area $1$ will always be more beneficial if spent before any time spent in Area $2$, where you risk being kicked out. Let's call $\Delta t_2$ the time interval that might be convenient to try spending in Area $2$. In such a time interval, Area $1$ would grant a food gain equal to $0.2 \, food/s \cdot \Delta t_2$. I don't like carrying numerical values, so I'll call $g_1 = 0.2 \,food/s$ and $g_2 = 1 \,food/s$, moreover $T = 600 \,s$. Now we have to evaluate the expected food gain in Area $2$ and maximize the difference.
I'm going to call $\tau$ the time spent in Area $2$, that is $\tau = t - \Delta t_1$. In Area $2$ there is a constant kick-out rate (probability per unit time), thus we have an exponential kick-out probability density function $f(\tau) = pe^{-p\tau}$. The expected survival time $E[\tau]$ ia evaluated as follows:
$$
\begin{align}
E[\tau]&=\int_0^{\Delta t_2}\tau \cdot p e^{-p\tau} d\tau + \Delta t_2 \cdot \int_{\Delta t_2}^\infty p e^{-p\tau}d\tau=\\
&=\left[\tau(-e^{-p\tau}) \right]_0^{\Delta t_2}-\int_0^{\Delta t_2}(-e^{-p\tau}) d\tau \;-\Delta t_2\cdot \left[e^{-p\tau} \right]_{\Delta t_2}^\infty=\\
&=-\Delta t_2 \, e^{-p\Delta t_2} \;- \left[\frac{1}{p}\cdot e^{-p\tau} \right]_0^{\Delta t_2} \; + \Delta t_2 \, e^{-p\Delta t_2}=\\
&=-\frac{1}{p}\cdot e^{-p\Delta t_2}+\frac{1}{p}=\\
&=\frac{1}{p}\left(1-e^{-p\Delta t_2}\right)
\end{align}
$$
Let's check for correct limits:
$$
\begin{align}
p\rightarrow 0 &\Rightarrow E[\tau]\rightarrow \Delta t_2\\
p\rightarrow \infty &\Rightarrow E[\tau]\rightarrow 0\\
\end{align}
$$
It's time to assess the food gain increment (note that $G_1$ is not the food gained during the first time interval, but what would be gained staying in Area $1$ during $\Delta t_2$):
$$
\begin{align}
\Delta G = G_2-G_1 &= g_2 \cdot \frac{1}{p}\left(1-e^{-p\Delta t_2}\right) - g_1 \cdot\Delta t_2\\
\frac{\Delta G\cdot p}{g_2} &= 1-e^{-p\Delta t_2} - \frac{g_1}{g_2} \cdot p\Delta t_2\\
y &= 1 - e^{-x} - \frac{g_1}{g_2} \cdot x\\
\end{align}
$$
where the last line follows a substitution of variables to underline the function that has to be maximized. Note that $y(0)=0$ and that $y(x)$ is positive up to a certain value of $x$, so that we are sure to encounter a maximum. So,
$$
y'= e^{-x}-\frac{g_1}{g_2} \\
$$
$$
\begin{align}
y'\geqslant 0 \iff& e^{-x}\geqslant\frac{g_1}{g_2}\\
&-x\geqslant \ln \frac{g_1}{g_2}\\
&\;\; x\leqslant \ln \frac{g_2}{g_1}\\
\end{align}
$$
Finally, $t^* = T-\Delta t_2 = T-\frac{1}{p} \ln \frac{g_2}{g_1}$ is the ideal time to move to Area $2$, provided that $g_2 \gt g_1$, otherwise it woudn't obviously be convenient to ever switch to Area $2$. At the opposite extreme, if $ \Delta t_2 \gt T$, the best strategy is to spend all the time in Area $2$.
With the actual values of the question, we have $t^* = 600 \, s - 1/p * \ln\frac{1}{0.2}$ and it would be beneficial to start in Area $2$ since the beginning whenever $p\leqslant 2.68 \cdot 10^{-3} \, s^{-1}$.
