Let's determine the maximum or minimum value of the function

$$
y=\left(x+\frac{a}{2}\right)^{2}-(3 x+2 a)^{2}
$$

knowing that $y$ takes its extreme value when

$$
x=-\frac{11}{8}
$$
The given function can also be written as:

$$
y=-8 x^{2}-11 a x-\frac{15}{4} a^{2}
$$

The coefficient of \(x^{2}\) is negative; therefore, the function has a maximum. This maximum value is attained when

$$
x=-\frac{11 a}{16}
$$

However, according to the condition,

$$
-\frac{11 a}{16}=-\frac{11}{8}
$$

and thus

$$
a=2
$$

Therefore, the maximum value of the function is:

$$
\frac{4 \cdot 8 \cdot 15-121 \cdot 4}{-4 \cdot 8}=\frac{1}{8}
$$