Let $x$ and $y$ be the two numbers.
$ x^{2} + y^{2} = 97 \quad \longrightarrow (1) $
The geometric mean cannot exceed the arithmetic mean. $ \boxed{ \text{AM} \geqslant \text{GM}} $
$ \Rightarrow \boxed{\frac{a_{1} + a_{2} + \dots + a_{n} } {n} \geqslant \sqrt[n]{a_{1} a_{2} \dots a_{n}}} $
Now$, \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{x^{2} \cdot y^{2}} $
$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant \sqrt{(xy)^{2}} $
$ \Rightarrow \frac{x^{2} + y^{2}}{2} \geqslant xy $
$ \Rightarrow x^{2} + y^{2} \geqslant 2xy $
$ \Rightarrow 97 \geqslant 2xy \quad [\because \text{From equation (1)}]$
$ \Rightarrow 2xy \leqslant 97 $
$ \Rightarrow xy \leqslant \frac{97}{2} $
$ \Rightarrow \boxed{xy \leqslant 48. 5}$
So$,xy$ cannot be more than $48. 5.$
$\therefore$ Only option $\text{(C)}$ not possible.
Correct Answer $: \text{C}$