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?

So you have a problem to compute the critical points of $f\mid_{S}$ where
$S=g^{-1}(4)$ with $g(x,y)=x^2+y^2$.
You need to solve the equation $\nabla f(x,y)=\lambda \nabla g(x,y)$ for some $\lambda\in\mathbb R$. You get
$$
\nabla f(x,y)=\lambda\nabla g(x,y)\Leftrightarrow \begin{pmatrix}1\\2\end{pmatrix}=\lambda\begin{pmatrix}2x\\2y\end{pmatrix}.
$$
You get $x=\frac1{2\lambda}$ and $y=\frac{1}{\lambda}$. Since $(x,y)\in S$ we conclude
$$
4=x^2+y^2=\frac1{4\lambda^2}+\frac1{\lambda^2}\Rightarrow 4\lambda^2=\frac54\Rightarrow \lambda^2=\frac5{16}\Rightarrow \lambda=\pm\frac{\sqrt{5}}4.
$$
Hence $f\mid_S$ has the critical points
$$
(x,y)\in\left\{\left(\frac{2}{\sqrt{5}},\frac{4}{\sqrt{5}}\right),\left(-\frac{2}{\sqrt{5}},-\frac{4}{\sqrt{5}}\right)\right\}
$$
Since $S$ is compact, one of theses critical points is the global minimum while the other is the global maximum. We compute
$$
f\left(\frac{2}{\sqrt{5}},\frac{4}{\sqrt{5}}\right)=\frac{10}{\sqrt{5}}=2\sqrt{5}
$$
and
$$
f\left(-\frac{2}{\sqrt{5}},-\frac{4}{\sqrt{5}}\right)=-\frac{10}{\sqrt{5}}=-2\sqrt{5}.
$$
These are the maximal and minimal values of $f$ on $x^2+y^2\leq 4$.
