Suppose a machine shows a particular reading (R) dependent on the weight (w) put on it, given by: $$R(w) = \frac{50e^w}{10+e^w}$$ Find the minimum possible value of r such that R(w)< r for all w, where r belongs to integers.
I am not getting any idea on how to proceed. The correct answer is 50
It looks like this problem is just asking for a maximum (or, more specifically, the upper bound of the function). Here's how we get 50:
$$y = \frac{50e^x}{10+e^x}$$ $$y = \frac{e^x}{e^x}\frac{50}{\frac{10}{e^x}+1}$$ $$y = \frac{50}{\frac{10}{e^x}+1}$$ $$\lim_{x\to\infty} \frac{50}{\frac{10}{e^x}+1} = 50$$
This works because $y$ is always increasing. We could also find the minimum by taking $lim_{x\to -\infty} \frac{50}{\frac{10}{e^x}+1}$, which yields 0. Therefore, $R = \{y\in\mathbb{R} \mid 0 < y < 50\}$ and $r=50$. A look at the graph confirms this.
