Why is the maximum likelihood estimator $y_{\operatorname{max}}$ instead of $y_{\operatorname{min}}$ Use the method of maximum likelihood to estimate the parameter $\theta$ in the uniform pdf,
$f_y(y ;\theta)=\frac{1}\theta$ where $0\leq y\leq\theta$
According to the solution manual $\theta_e=y_{\operatorname{max}}$, however the likelihood function $L(\theta)=(\frac{1}\theta)^n,$
For me, what would maximize $L(\theta)$ would be $\theta_e=y_{\operatorname{min}}$

If $\theta$ is less than any of your data points then the value of your product likelihood will be $0$, because:
$$f_y(y,\theta) = \left\{\begin{array}{ll}
\frac{1}{\theta} & \mathrm{\ for\ } 0 \le y \le \theta \\
0 & \mathrm{\ otherwise.}
\end{array}\right.$$ So, for a given value of $\theta$, the product likelihood will be $\theta^{-m} 0^{N-m}$ where $N$ is the number of data points and $m$ is the number of data points, $y_i$ that satisfy $y_i \le \theta$. 
