The variables $p,q,r$ and $s$ are correlated with each other with the following
relationships $\dfrac{s^{0.5}}{p}=\dfrac{q}{r^2}$ .The ranges of values of
$p,q$ and $r$ are respectively: $-0.04\leq p\leq -0.03,-0.25\leq q\leq-0.09,1\leq r\leq 7$.
Determine the difference between the maximum and minimum value of $s$?
$a.)\ -0.02\\ b.)\ 0.02\\ c.)\ 0.1 \\ d.)\ \text{none of these}$
I concluded that the answer is $d.)$ but the book is saying something else. So I am not sure if I am missing something.
Solution by book
The equation can be rewritten as $s^{0.5}=\dfrac{pq}{r^2}$.
The maximum of $s^{2}$ will happen when we take exterme
values of $p,q$ and $r$ in order to support the maximization of $s^{2}$
Taking $p=-0.4,\ q=-0.25$
and $r=1$ we get $s^2=\dfrac{-0.04\times -0.25}{1}\rightarrow s^{0.5}=1$
Maximum value of $s=0.01$
Minimum value of $s=-0.01$
So, required difference $=0.02$
Since we have $\sqrt{s} = \frac{pq}{r^2}$ then the maximum value of $s$ will be given by maximising $p^2q^2$ and minimising $r^4$.
Then, maximising $p^2q^2$ and minimising $r^4$ yields $$\max s = 0.04^2 \cdot 0.25^2 = 0.0001$$
To minimise $s$, we need to maximise $r^4$ by taking $r=7$ and minimise $p^2q^2$ so that we get $$\min s = \frac{0.03^2 \cdot 0.09^2}{7^4}.$$
Hence the difference is given by $$\max s - \min s = 0.0001 - \frac{0.00000729}{2401}$$