Disease virulence The Kermack-Mckendrick model for infectious disease transmission can be used to predict the population size $$ P $$ as a function of the disease's virulence, Virulence measures the deadliness of a disease. The population size $$ P $$ is high when virulence v is low because it is not very deadly, Surprisingly $$ P $$ is also high when $$ v $$ is high because then the disease is so deadly it kils people before they can pass it on, so fewer people actually get infected. Suppose, $$ P(v)=\frac{10+v+v^{2}}{1+v} $$ for values of v from 0 to 10 inclusive, What virulence values produce the largest and smallest population sizes? What are those populations?

Question

UpStudy AI · Accepted Answer

We are given
$$
P(v)=\frac{10+v+v^2}{1+v},\quad 0\le v\le 10.
$$
A useful first step is to perform polynomial division on the numerator by $$1+v$$. Dividing $$v^2+v+10$$ by $$v+1$$ gives

$$
v^2+v+10 = (v+1)\cdot v + 10,
$$
so

$$
P(v)= v + \frac{10}{v+1}.
$$

### Step 1. Differentiate to Find Critical Points

Differentiate $$P(v)$$ with respect to $$v$$:
$$
P(v)= v + \frac{10}{v+1}.
$$
The derivative is
$$
P'(v)= 1 - \frac{10}{(v+1)^2}.
$$
Set $$P'(v)=0$$ to find the critical points:
$$
1-\frac{10}{(v+1)^2}=0\quad\Longrightarrow\quad\frac{10}{(v+1)^2}=1\quad\Longrightarrow\quad (v+1)^2=10.
$$
Taking the positive square root (since $$v+1>0$$) gives
$$
v+1=\sqrt{10}\quad\Longrightarrow\quad v=\sqrt{10}-1.
$$
Since $$\sqrt{10}\approx 3.16$$, we have $$v\approx 2.16$$.

### Step 2. Verify if $$v=\sqrt{10}-1$$ is a Minimum or Maximum

Differentiate $$P'(v)$$ to find the second derivative if needed:
$$
P''(v)= \frac{20}{(v+1)^3}.
$$
Since $$(v+1)^3>0$$ for $$ v \ge 0$$, the second derivative is positive, confirming that $$v=\sqrt{10}-1$$ is a local minimum.

### Step 3. Evaluate $$P(v)$$ at the Endpoints and the Critical Point

- **At $$v=0$$:**
  $$
  P(0)= 0+\frac{10}{0+1}= 10.
  $$
  
- **At $$v=10$$:**
  $$
  P(10)= 10+\frac{10}{10+1}= 10+\frac{10}{11}\approx10+0.9091= 10.9091.
  $$
  
- **At $$v=\sqrt{10}-1$$:**
  $$
  P(\sqrt{10}-1)= (\sqrt{10}-1)+\frac{10}{\sqrt{10}}.
  $$
  Using $$\sqrt{10}\approx3.16$$:
  $$
  P(\sqrt{10}-1)\approx (3.16-1)+\frac{10}{3.16}\approx 2.16+3.16= 5.32.
  $$

### Conclusion

- The **smallest population** occurs at $$v=\sqrt{10}-1$$ with
  $$
  P(\sqrt{10}-1)\approx 5.32.
  $$
- The **largest population** occurs at $$v=10$$ with
  $$
  P(10)\approx 10.91.
  $$
The smallest population occurs when virulence $$ v $$ is approximately $$ \sqrt{10} - 1 $$ (about 2.16), with a population size of about 5.32. The largest population occurs when $$ v = 10 $$, with a population size of approximately 10.91.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Calculus Questions