Given the demand equation $$p=\frac{240}{q+15}$$ Where $15 \leq q \leq 105$, for what value of $q$ is $|\eta|$ a maximum? for what value is it a minimum? Well, i try the following. Note that:
$$\eta=\frac{dq}{dp} \frac{p}{q}$$
As $p=\frac{240}{q+15}$, then $q=\frac{240-15p}{p}$ and $\frac{dq}{dp}=\left(\frac{240-15p}{p}\right)'=\frac{\left(240-15p\right)'\:p-p'\left(240-15p\right)}{p^2}=-\frac{240}{p^2}$ and
$$\eta=-\frac{240}{p^2} \frac{240}{\frac{q+15}{\frac{240-15p}{p}}}=-\frac{57600\left(240-15p\right)}{p^3\left(q+15\right)}$$
Now, I was thinking of evaluating the behavior of the derivative of this function to try to find the maximum and minimum, but it is a function that depends on both $p$ and $q$ so it is not clear to me what I can do from that $\eta$ expression to find the maximum and minimum. Any suggestions?
The derivative is right. Now we insert the terms into the formula, $p$ remains $p$.
$$\eta=-\frac{240}{p^2}\cdot \frac{p}{\frac{240-15p}{p}}=-\frac{240}{p^2}\cdot \frac{p^2}{240-15p}=-\frac{240}{240-15p}$$
$$=\frac{240}{15p-240}=\frac{16}{p-16}$$
Next we have to evaluate the limits of $p$. For $q=15$ we obtain the upper bound $p^o=\frac{240}{15+15}=8$. For $q=105$ we obtain the lower bound $p^u=\frac{240}{105+15}=\frac12$. The we have $\frac12\leq p \leq 8$. Now you can maximize and minimize $\eta$ with the regard to the bounds.