A worker is disposing of radioactive material of mass $\mu$ and needs to minimize her exposure. Being near the radioactive material exposures her to radiation at a rate of $\frac {dE_n}{dt}=c\mu$, where $E_n$ is the radiation she gets from being near the material and $c$ is a constant. However having to pick up the material gives her some addition exposure at a rate of $\frac{dE_p}{dt}=k(\mu')^2$, where $E_p$ is the exposure due to picking up the material, $\mu'$ is the time derivative of $\mu$, and $k$ is another constant.
$1)$ Find a functional for total exposure over some time $\Delta t$.
$2)$ Minimize that functional to find a DE for the amount of material $\mu(t)$.
Here's what I've done:
$1)$ $E_{total} = \int_0^{\Delta t} \frac {dE}{dt} dt = \int_0^{\Delta t} \frac {dE_n}{dt} + \frac {dE_p}{dt}dt = \int_0^{\Delta t}[c\mu(t)+k(\mu'(t))^2]dt$. That one seems relatively easy so I think I did it right.
$2)$ Here I'm having some problems because doing it two different ways, I get two different answers.
First way: use the characteristic equation $\frac {\partial F}{\partial y} - \frac {d}{dx}\frac{\partial F}{\partial y'} =0$.
So in this case we have $c-\frac{d}{dt}(2k\mu'(t))=0$. Therefore $\mu'' = \frac {c}{2k}$. Therefore $\mu(t) = \frac {c}{2k}t^2+\alpha t+\beta$, for some constants $\alpha$ and $\beta$ which we find given some initial conditions.
The other way is to use the fact that $F$ doesn't depend on $t$ explicitly (only $\mu$ and $\mu'$ appear explicitly in the equation) so the characteristic equation should "simplify" to $F-\mu'\frac{\partial F}{\partial \mu'}=\gamma$ for some constant $\gamma$.
Then $c\mu+k(\mu')^2-2k(\mu')^2=c\mu-k(\mu')^2=\gamma$. Since this is nonlinear, I just plugged it into WolframAlpha and am getting some complicated thing with product log functions -- which doesn't seem to be the same as the simple quadratic I got above.
So my questions: Did I do part $1$ correctly? If so, why am I getting different answers using the two different approachs to part $2$?
You did part (1) correctly, and your Euler-Lagrange equation $$ \mu'' = \frac{c}{2k} $$ is correct. This gives a solution $ \mu(t) = \frac{c}{4k} t^2 + \alpha t + \beta t $ (which is off by a factor of 2 what you wrote). This also satisfies the nonlinear equation. To see this, we calculate
\begin{align*} c\mu - k(\mu')^2 &= \frac{c^2}{4k}t^2 + c\alpha t + c\beta - k(\frac{c}{2k}t + \alpha)^2 \\&= \frac{c^2}{4k}t^2 + c\alpha t + c\beta - \frac{c^2}{4k} t^2 - c\alpha t - k\alpha^2 \\&= c\beta - k\alpha^2. \end{align*}
The right-hand side is independent of time, so it satisfies the equation you wrote.
In summary, you did everything right except miss a factor of 2 in your quadratic solution. I don't know what Mathematica is doing with the log-products, maybe it's confused because of the undetermined constant in the ODE.