For some real numbers $a$ and $b$, the system of equations $x+y=4$ and $(a+5)x+(b^2-15)y=8b$ has infinitely many solutions for $x$ and $y$. Then, the maximum possible value of $ab$ is

The correct option is 33.

We are given the following system of linear equations in variables $x$ and $y$:
1) $x+y=4$
2) $(a+5)x+(b^2-15)y=8b$

For a system of two linear equations of the form $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$ to have infinitely many solutions, the coefficients of the two equations must be proportional. That is:

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Applying this condition to our system of equations, we get:

$\frac{a+5}{1}=\frac{b^2-15}{1}=\frac{8b}{4}$

Simplifying the ratios, we obtain two equations:

$a+5=b^2-15$

$a+5=2b$

From the second equation, we can express $a$ in terms of $b$:

$a=2b-5$

Now, we equate the second and third parts of our ratio equation:

$b^2-15=2b$

Rearranging this into a standard quadratic equation in terms of $b$ gives:

$b^2-2b-15=0$

We can factor this quadratic equation as follows:

$(b-5)(b+3)=0$

This gives two possible values for $b$:
Case 1: $b=5$
Case 2: $b=-3$

Next, we determine the corresponding values of $a$ and calculate the product $ab$ for each case:

For Case 1 ($b=5$):
$a=2\left(5\right)-5=5$
Thus, the product is:
$ab=5\times 5=25$

For Case 2 ($b=-3$):
$a=2(-3)-5=-11$
Thus, the product is:
$ab=(-11)\times (-3)=33$

Comparing the values of $ab$ from both cases, the maximum possible value is 33.