Express $y = 4x^2 + 24x + 49$ in the form $a(x+b)^2 + c$. Hence, state the maximum value of $\frac{1}{y}$ and the corresponding value of $x$ at which $\frac{1}{y}$ is maximum.

1. Factor out the coefficient of $x^2$ from the $x$ terms

$y = 4(x^2 + 6x) + 49$

2. Complete the square inside the bracket and simplify

$y = 4\left[ (x + 3)^2 - 3^2 \right] + 49$
$y = 4(x + 3)^2 - 36 + 49$
$y = 4(x + 3)^2 + 13$

1. Determine the completed square form of $y$ from part (i)

From part (i), $y = 4(x^2 + 6x) + 49 = 4(x+3)^2 - 36 + 49 = 4(x+3)^2 + 13$.

2. Find the minimum value of $y$ and the corresponding $x$

Since $4(x+3)^2 \geqslant 0$ for all real $x$, the minimum value of $y$ is $13$, which occurs when $x + 3 = 0 \implies x = -3$.

3. Determine the maximum value of $\frac{1}{y}$

The expression $\frac{1}{y}$ is maximum when $y$ is positive and at its minimum.
Therefore, the maximum value of $\frac{1}{y}$ is $\frac{1}{13}$, occurring at $x = -3$.