Suppose $X_1, \dots, X_n \overset{\text{iid}}{\sim}\dfrac{x}{\theta}\exp\left(-\dfrac{x^2}{2\theta}\right)\mathbf{1}_{(0, \infty)}(x)$, $\theta > 0$. At the end of the day, my goal is to calculate $$\max_{\theta > 16}L(\theta)$$ where $L$ is the likelihood function, i.e., $$L(\theta) = \dfrac{\prod_{i=1}^{n}x_i}{\theta^n}\exp\left(\dfrac{-1}{2\theta}\sum_{i=1}^{n}x_i^2\right)\text{.}$$ In order to do this, we have to find $\theta$ such that $L$ is maximized. Using the typical methods, I find that $$\hat{\theta}_n = \dfrac{\sum_{i=1}^{n}X_i^2}{2n}$$ is the maximum likelihood estimator of $\theta$, assuming $\theta > 0$.
However, when restricting $\theta$ to $\{\theta: \theta > 16\}$, the solution I have says that the maximum likelihood estimator is $$\hat{\theta}_n = \max\left(16, \dfrac{\sum_{i=1}^{n}X_i^2}{2n}\right)$$ is the MLE. Why is this? Intuitively, this doesn't make any sense to me because one would think that one wants $\hat\theta_n$ to be as small as possible, since increasing $\theta$ would decrease $L$, holding the $\{x_i\}$ constant. So I think $\max$ should be replaced with $\min$ in the equation for $\hat\theta_n$ above. However, yet at the same time, we are restricting $\theta > 16$.
Could someone please provide me with some insight for this?
$L(\theta)$ goes up when $\theta$ goes from $0$ to the typical MLE point (the first $\hat\theta$ you get), reaches peak at $\hat\theta$ and goes down henceforth. This should give you the idea.