This is an extremely basic question, but I would appreciate some clarification. Consider the following problem. Fix a continuously differentiable function $F(x,y,y')$, $-\infty < a < b < \infty$, and $A, B \in \mathbb{R}$. Our problem is the standard problem of finding a function $y = f(x)$ such that the value of $$J(f) = \int_{a}^{b}F(x,y,y')dx$$ at $y = f(x)$ is extremal (locally), and $F(a) = A$, $F(b) = B$. Suppose $F$ is of the form $F(x,y,y') = H(x)y$. We know that necessary conditions for a (local) extremum are $$\tag{$1$}\label{1}0 = \frac{\partial}{\partial{y}}F - \frac{d}{dx}\frac{\partial}{\partial{y'}}F$$ However, given the form of $F$, we have $\frac{\partial}{\partial{y}}F = H(x)$ and $\frac{\partial}{\partial{y'}}F = 0$.
Now, if this were a simple calculus question, and we were asked to choose say $x \in [a,b]$ to maximize $g(x)$ where $g(x) = kx$ for some constant $k$, we'd glance at it, see that it was linear and know that a corner $x$ was the maximizer.
Hence, the problem which results in expression \ref{1} seems like the variational analogue of the basic calculus problem, and it seems intuitive that again there should be some sort of "corner" solution. Is this the correct way to think about it? If so, what precisely is a corner solution in these sorts of problems? Finally, how might things change should we allow $a$ and $b$ to be $-\infty$ and $\infty$, respectively?
It might be helpful to stress that we have to solve wrt. $f$; not wrt. $x$, if that was ever in doubt. The result is:
If $H$ is the zero function (a.e.), then OP's functional $J[f]$ is the zero functional, and any function $f$ is stationary.
If $H$ is not the zero function (a.e.), then OP's functional $J[f]$ has no stationary function $f$. More details: Define preimage sets $$S_{\pm}~:=~ H^{-1}(\mathbb{R}_{\pm}) ~\subseteq~ [a ,b]. $$ Let $1_{S_{\pm}}$ be the indicator function. Define $ \pm\infty \cdot0 :=0$. The would-be maximum/minimum functions are then played by $f=\pm \infty \cdot 1_{S_{\pm}}$, respectively.