Disease virulence The Kermack-Mckendrick model for infectious disease transmission
can be used to predict the population size as a function of the disease’s virulence,
Virulence measures the deadliness of a disease. The population size is high when
virulence v is low because it is not very deadly, Surprisingly is also high when 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,

for values of v from 0 to 10 inclusive, What virulence values produce the largest and
smallest population sizes? What are those populations?

Ask by Whittaker Barber. in Canada

Mar 19,2025

Question

Disease virulence The Kermack-Mckendrick model for infectious disease transmission
can be used to predict the population size as a function of the disease’s virulence,
Virulence measures the deadliness of a disease. The population size is high when
virulence v is low because it is not very deadly, Surprisingly is also high when 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,

for values of v from 0 to 10 inclusive, What virulence values produce the largest and
smallest population sizes? What are those populations?

Ask by Whittaker Barber. in Canada

Mar 19,2025

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.

Solución de inteligencia artificial de Upstudy

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Solución

Step 1. Differentiate to Find Critical Points

Step 2. Verify if is a Minimum or Maximum

Step 3. Evaluate at the Endpoints and the Critical Point

Conclusion

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