What is the maximum possible value of a + 4b? Let $a$ and $b$ be real numbers so that the equation $x^2 − (a + b)x + 9(a^2 + b^2 ) = \frac{17}{18}$ has an integer solution. What is the maximum possible value of $a + 4b$?
I can't figure out if I need to use the AM-GM inequality or the quadratic equation. Any help would be appreciated. 

Let $c = a + 4b$. Then $a = c - 4b$ and the equation becomes:
$x^2 - (c - 4b + b)x + 9(c^2 - 8bc + 16b^2 + b^2) = x^2 + (3b - c)x + 9(17b^2 - 8bc + c^2) = \frac{17}{18}$
Rearranging, we get 
$cx + 9c^2 - 8bc = -x^2 + 3bx - 17b^2 - \frac{17}{18}$
or 
$f(c) = 9c^2 + (x - 8b)c = -x^2 + 3bx - 17b^2 - \frac{17}{18}$
A critical point of $f(c)$ occurs when the derivative of $f(c)$ with respect to $c$ is equal to zero (the right side doesn't matter because all constants with respect to the variable $c$ become zero with the derivative):
$\partial_c f(c) = \partial_c((9c^2 + (x - 8b)c) = 18c + (x - 8b) = 0$
$c = \frac{x - 8b}{18}$
Because the integer variable $x$ is not bounded above and the real number variable $b$ is not bounded below, the critical point $c$ has no maximum over the entire range of values for $x$ and $b$, regardless if it is a maximum, a minimum, or a saddle point with fixed $x$ and $b$. Since $c = a + 4b$, there is no maximum for $a + 4b$ either. 
