Application of Calculus to medicine I was wondering if there's a function or a derivative to model the following problem (since this problem is from a Calculus book) or it's solved just using the Rule of three with percentages

 A: Ok, let's begin with the first problem.
1. According to the info:
$$ \dfrac{dH}{dt} = kH, $$
with $H$ the volume of red blood cells. We also have:
$$H(t=0)=5*0.45=2.25\, \mathrm L, \quad \text{$t$ in hours}.$$
At the end, the patient has lost $2.5\, \mathrm L$ of blood, which means that the patient has $2.5\,\mathrm L$ of blood (remember that before the surgery the patient has $5\, \mathrm L$ of blood). In order to know $H(t= 4)$, we do the following:
$$2.5 * 0.45 = 1.125\, \mathrm L.$$
$1.125\, \mathrm L$ is the volume of red blood cells after $4$ hours.
2. For the second problem, a percentage of blood is extracted before the surgery. We denote it like this:
$$\text{blood extracted} = 5x, \quad \text{with $0\leq x \leq 1$}.$$
The remaining blood is:
$$ 5 - 5x,$$
and the volume of red blood cells is:
$$ 0.45(5 - 5x).$$
After $4$ hours, the patient has lost $2.5\, \mathrm L$, which means $5-5x-2.5$, so the volume of red blood cells he has lost is:
$$0.45(5-5x-2.5).$$
The concentration can never be allowed to drop below $25 \,\%$, therefore, the maximum amount of blood that can be extracted occurs when he has a concentration of red blood cells $0.25(5 -5 x)$ after $4$ hours. Putting all of this into an equation:
$$ 0.45(5 - 5x) - 0.45(5-5x-2.5) = 0.25(5 -5 x).$$
And $x=0.1$, the maximum amount of blood that can be extracted is $10\,\%$.
3. With this information, can you do the third problem?
