Let $X_1,...,X_n$ random sample of $X$~$N(\mu,\sigma^2)$ with known $\sigma^2$.Take $a=.05$ find the expression for power function of the likelihood-ratio test $$H_0:\mu\leq 0\space vs\space H_1:\mu>0$$
I know that $\hat{\mu}=\overline{X}$
$$\lambda(x)=\frac{\sup_{\Theta_0}L(\theta|x)}{\sup_{\Theta}L(\theta|x)}=\frac{(2\pi)^\frac{-n}{2}(\sigma^2)^{\frac{-n}{2}}e^{-\frac{1}{2\sigma^2}\sum (x_i-\mu_0)^2}}{(2\pi)^\frac{-n}{2}(\sigma^2)^{\frac{-n}{2}}e^{-\frac{1}{2\sigma^2}\sum (x_i-\overline{x})^2}}$$
$$\sum(x_i-\mu_0)^2=\sum(x_i-\overline{x})^2+n(\overline{x}-\mu_0)^2$$ then $$\lambda(x)=e^{-\frac{n}{2\sigma^2}(\overline{x}-\mu_0)^2}$$ so $$\lambda(x)\leq c\Leftrightarrow -\frac{n}{2\sigma^2}(\overline{x}-\mu_0)^2\leq log(c)\Leftrightarrow \overline{x}\geq-\sigma\sqrt{\frac{2log(c)}{n}}+\mu_0$$
In the answer they define the critical region as $\overline{X}\geq c\frac{\sigma}{\sqrt{n}}$.
But I don't know how to get it, perhaps they disregarded some terms to facilitate the calculation of the power function