MLE of $\mu$ given $X_1\sim \mathsf N(\mu, 4)$ and $X_2\sim \mathsf{N}(\mu, 16)$ 
Let $X_1\sim \mathsf N(\mu, 4)$ and $X_2\sim \mathsf{N}(\mu, 16)$
  where $X_1$ and $X_2$ are independent.  Find the maximum likelihood
  estimator $\hat{\mu}$ of $\mu$ if it exists.

We have
$$\begin{align*}
L(\mu\mid x_1, x_2)
&=\frac{1}{\sqrt{8\pi}}e^{-\frac{(x_1-\mu)^2}{8}}\cdot \frac{1}{\sqrt{32\pi}}e^{-\frac{(x_2-\mu)^2}{32}}\\\\
&=\frac{1}{\sqrt{8\pi}\sqrt{32\pi}}e^{-\frac{(x_1-\mu)^2}{8}-\frac{(x_2-\mu)^2}{32}}
\end{align*}$$
Then
$$logL(\mu\mid x_1, x_2)=log\left(\frac{1}{\sqrt{8\pi}\sqrt{32\pi}}\right)-\frac{(x_1-\mu)^2}{8}-\frac{(x_2-\mu)^2}{32}$$
Then we set
$$\frac{\partial log(L)}{\partial\mu}=\frac{1}{4}(x_1-\mu)+\frac{1}{16}(x_2-\mu)=0$$
Solving for $\mu$ I get that $\hat{\mu}=\frac{4}{5}x_1+\frac{1}{5}x_2$
This doesn't satisfy my intuition. I would have thought $\hat{\mu}=\frac{x_1+x_2}{2}$
Checking that this is in fact a maximum, I get that
$$\frac{\partial}{\partial\mu}\left(\frac{1}{4}(x_1-\mu)+\frac{1}{16}(x_2-\mu)\right)=-\frac{5}{16}\lt0\text{ } \checkmark$$
Is this a valid solution?
Edit:
I think the MLE makes sense actually. It gives more weight to the random variable with a lower variance.
 A: The procedure is not right in my opinion. Note that you have a sample size of $n$. The likelihood function is
$$L(\mu | x_{11},x_{12},\ldots,x_{1n},x_{21},x_{22},\ldots,x_{2n})=\prod_{i=1}^n \left( \frac1{16\cdot \Pi}\right)\cdot \Large{e^{-\frac{(x_{1i}-\mu)^2}{8}+\frac{(x_{2i}-\mu)^2}{32}}}$$
$$=\left( \frac1{16\cdot \Pi}\right)^n\cdot \prod_{i=1}^n  \Large{e^{-\frac{(x_{1i}-\mu)^2}{8}+\frac{(x_{2i}-\mu)^2}{32}}}$$
Taking logs
$$-n\cdot \ln (16\cdot \Pi)-\sum_{i=1}^n\left( \frac{(x_{1i}-\mu)^2}{8}+\frac{(x_{2i}-\mu)^2}{32} \right)$$
Taking derivative w.r.t. $\mu$ and set it equal to $0$
$$\sum_{i=1}^n\left( 2\cdot \frac{x_{1i}-\mu}{8}+2\cdot \frac{x_{2i}-\mu}{32} \right)=0$$
$$\frac14\sum_{i=1}x_{1i}+\frac1{16}\sum_{i=1}x_{2i}=\frac5{16}\cdot n\cdot \mu$$
$$\frac4{16}\sum_{i=1}x_{1i}+\frac1{16}\sum_{i=1}x_{2i}=\frac5{16}\cdot n\cdot \mu \quad |\cdot 16$$
$$4\sum_{i=1}x_{1i}+\sum_{i=1}x_{2i}=5\cdot n\cdot \mu \quad $$
Then we use $\frac1n \sum\limits_{i=1}^nx_{1i}=\overline{x_1} \Rightarrow \sum\limits_{i=1}^nx_{1i}=n\cdot \overline{x_1}$ Similar for the sample of $X_2$
$$4n\cdot \overline{x_1}+n\cdot \overline{x_2}=5\cdot n\cdot \mu \qquad |:n$$
$$4\cdot \overline{x_1}+ \overline{x_2}=5\cdot  \mu $$
Dividing the equation by 5 we obtain the MLE for $\mu$
$$\boxed{\hat \mu=\frac{4\cdot \overline{x_1}+ \overline{x_2}}{5}}$$
So you take a sample of $n$ from both distributions each and  with the sample means, $\overline{x_1}$ and $\overline{x_2}$, you can estimate $\mu$.
