Question: If

$$S\_S \ge \frac{120x}{\frac{60}{x}+1}$$

and

$2x\le 45$

then what can be said about the range of $S\_S$?

Answer: 
For positive $x$, the numerator is a growing function and the denominator a decreasing one, so that the ratio is growing.

Hence,

$$\frac{120\cdot0}{\frac{60}{0}+1}\le S\_S\le\frac{120\frac{45}2}{60\frac2{45}+1}.$$

(Though the lower bound is meaningless as written, the limit is $0$.)
See [this](https://www.desmos.com/calculator/5dx7u9pqil).

$$S\_s(a)>S\_s(b) \space \text{for} \space a,b>0, a>b.$$

Hence the max on the positive is $S\_s(22.5)=\frac{8100}{11}$

We see $x=0$ cannot exist due to $\frac{60}{x}$ on the denominator.

Hence $$0<S\_s\le\frac{8100}{11}$$