Out for a hike

Let $M(d)$ be the amount of food in the backpack after having walked distance $d$ . Then we have $M(0)=50$ and $M^\prime(d) = -\gamma_0\big(1+\frac{M(d)}{W_{\text{body}}}\big)$ . Solving this ODE gives us $M(d)=210 e^{-\frac{d}{1920}}-160$ . Solving for $M(d)=0$ (in the reals), yields $d=1920\log\frac{21}{16}\approx522.113$ .

Food depletion is almost linear (compare to upper line $50-d/12$ ):

First I will find amount of food $dm$ needed to travel distance $dl$ with $m$ food in backpack.

$\displaystyle dm=\gamma_0 dl (1+\frac{m}{W_{body}})$

This can be easily transformed to the following integral

$\displaystyle dl=\frac{dm}{\gamma_0 (1+\frac{m}{W_{body}})}$

$\displaystyle dl=\frac{W_{body}}{\gamma_0}\frac{dm}{ (W_{body}+m)}$

$\displaystyle \int_{0}^{L}{dl}=\frac{W_{body}}{\gamma_0} \int_{0}^{M}{ \frac{dm}{ (W_{body}+m)}}$

And by integrating this we get:

$\displaystyle L= \frac{ W_{body} }{\gamma_0} \ln { \frac{M+W_{body}}{ W_{body}} }\approx 522.11 \text{ miles}$