Let $X_1,\dots,X_n\sim N(\mu,\sigma^2)$, $\sigma^2$ is unknow. We want to test
$H_0:\mu=5$ vs $H_1:\mu>5 $
$$f_{\theta}(\vec{x})=\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^n\exp\left[-\sum_{i=1}^n\frac{(x_i-\mu)^2}{2\sigma^2}\right]$$
$\underset{\sigma^2\geq 0 ,\;\mu\in\theta}{\sup}f_\theta(\vec{x})$
$$\ln f_\theta(\vec{x})=-\frac{1}{2\sigma^2}\left(\sum_{i=1}^n(x_i-\mu)^2\right)-\frac{n\ln 2\pi\sigma^2}{2}$$
$$\frac{\partial\ln f_\theta(\vec{x})}{\partial \mu}=\frac{1}{\sigma^2}\sum_{i=1}^n(x_i-\mu))=\frac{n(\bar{x}-\mu)}{\sigma^2}$$ Then
$$\frac{\partial\ln f_\theta(\vec{x})}{\partial\mu}=0\Leftrightarrow \frac{n(\bar{x}-\mu)}{\sigma^2}=0\Leftrightarrow \mu=\bar{x}\rightarrow \hat{\mu}=\bar{x}\hspace{0.25cm}\text{max or min?}$$
Because $$\frac{\partial^2\ln f_\theta(\vec{x})}{\partial\mu^2}=-\frac{n}{\sigma^2}<0\hspace{0.5cm}\hat{\mu}\;\text{is a max}$$ Taking $\mu=\bar{x}$ we have $$\frac{\partial\ln f_\theta(\vec{x})}{\partial\sigma^2}=-\frac{n}{2\sigma^4}\left(\sigma^2-\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2\right)$$ Then $$\frac{\ln f_\theta(\vec{x})}{\sigma^2}=0\Leftrightarrow \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2\rightarrow\hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2\hspace{0.25cm}\text{max or min?}$$ Because $$\left.\frac{\partial^2\ln f_\theta(\vec{x})}{\partial (\sigma^2)^2}\right|_{\sigma^2=\hat{\sigma}^2}=-\frac{n^3}{2}\left(\sum_{i=1}^n(x_i-\bar{x})\right)^{-2}<0\hspace{0.5cm} \hat{\sigma}^2 \hspace{0.25cm}\text{it's a max}$$ Then $$\underset{\sigma^2\geq 0 ,\;\mu\in\theta}{\sup}f_\theta(\vec{x})=\left(\frac{1}{\sqrt{2\pi\hat{\sigma}^2}}\right)^n\exp\left[-\sum_{i=1}^n\frac{(x_i-\bar{x})^2}{2\hat{\sigma}^2}\right]=\left(\frac{1}{\sqrt{2\pi\hat{\sigma}^2}}\right)^ne^{-\frac{n}{2}}$$
$$\underset{\sigma^2\geq 0 ,\;\mu\in\theta_0}{\sup}f_\theta(\vec{x})$$ $$\hat{\sigma}_0^2=\frac{1}{n}\sum_{i=1}^n(x_i-\mu_0)^2$$ Then $$\underset{\sigma^2\geq 0 ,\;\mu\in\theta_0}{\sup}f_\theta(\vec{x})=\left(\frac{1}{\sqrt{2\pi\hat{\sigma}_0^2}}\right)^n\exp\left[-\sum_{i=1}^n\frac{(x_i-\mu_0)^2}{2\hat{\sigma}_0^2}\right]=\left(\frac{1}{\sqrt{2\pi\hat{\sigma}_0^2}}\right)^ne^{-\frac{n}{2}}$$
So $$\Lambda=\frac{\underset{\sigma^2\geq 0 ,\;\mu\in\theta_0}{\sup}f_\theta(\vec{x})}{\underset{\sigma^2\geq 0 ,\;\mu\in\theta}{\sup}f_\theta(\vec{x})}=\left(\frac{\hat{\sigma}^2}{\hat{\sigma}_0^2}\right)^{n/2}$$