Maximize linear function over disk of radius $2$ 
Maximize $f(x, y) = x + 2y$ with constraint $x^2 + y^2 \le 4$.

$f(x, y)$ has no CP's so thats something gone.
I considered the boundary/edge of $g(x, y) = x^2 + y^2$. And I got the point $(0, 2), (2, 0)$, which gave $f(0, 2) = f(2, 0) = 4$. However, the point $(2/\sqrt(5), 4/\sqrt(5))$ gives a larger max.
How do I approach this to get this point?

$$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 \leq 4\end{array}$$
Since the objective function is linear and nonzero, its gradient never vanishes. Thus, the maximum is attained at the boundary of the feasible region, i.e., on the circle of radius $2$ centered at the origin
$$\begin{array}{ll} \text{maximize} & x + 2y\\ \text{subject to} & x^2 + y^2 = 4\end{array}$$
Using polar coordinates instead, the objective function is
$$2 \cos (\theta) + 4 \sin (\theta) = 2 \sqrt{5} \left( \dfrac{2}{2 \sqrt{5}} \, \cos (\theta) + \dfrac{4}{2 \sqrt{5}} \, \sin (\theta) \right) = 2 \sqrt{5} \, \cos \left( \theta - \arctan (2) \right)$$
Thus, the maximum is $\color{blue}{2 \sqrt{5}}$, which is attained at $\theta = \arctan (2)$, i.e., at
$$x = 2\,\cos \left( \arctan (2) \right) = \dfrac{4}{2 \sqrt{5}} = \color{blue}{\dfrac{2}{\sqrt{5}}}$$
$$y = 2\,\sin \left( \arctan (2) \right) = \dfrac{8}{2 \sqrt{5}} = \color{blue}{\dfrac{4}{\sqrt{5}}}$$
