Given:
Some real numbers a and b, the system of equations x + y = 4 and (a + 5) x + (b2 - 15) y = 8b has infinitely many solutions for x and y.
Therefore, we can write it as:
\(⇒\) \(\frac{a+5}{1}=\frac{b^2-15}{1}=\frac{8b}{4}\)
Above equation can be used to find the value of a and b.
Let's get the value of b:
\(⇒\) \(\frac{b^2-15}{1}=\frac{8b}{4}\)
\(⇒\) b2 - 2b - 15 = 0
Therefore, the values of b can be 5 and -3 respectively.
Now, the value of a can be stated in terms of b, which is:
a + 5b = b2 - 15
\(⇒\) a = b2 - 20
So, when b = 5, then a = 52 - 20 = 5
And when b = -3, then a = 32 - 20 = -11
Now, the maximum value of ab is:
= (-3) × (-11)
= 33
So, the correct option is (C): 33.