Chapter 6.5, Problem 41E

a.

To determine

Find the maximum and minimum number of wolves.

Answer to Problem 41E

  $3000,1000$

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the given equation,

  $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$

The range of the sine function is, $\left(-1,1\right)$

So the wolves population is maximum when $\sin \left(\frac{\pi t}{6}\right)=1$ and minimum when $\sin \left(\frac{\pi t}{6}\right)=-1$ ,

  $\begin{array}{l}W=2000+1000\left(1\right)\\ W=3000\\ W=2000+1000\left(-1\right)\\ W=1000\end{array}$

Hence, the maximum and minimum number of wolves are $3000,1000$ .

b.

To determine

Find the maximum and minimum number of sheep.

Answer to Problem 41E

  $15,000,5000$

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the given equation,

  $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$

The range of the cosine function is, $\left(-1,1\right)$

So the wolves population is maximum when $\cos \left(\frac{\pi t}{6}\right)=1$ and minimum when $\cos \left(\frac{\pi t}{6}\right)=-1$ ,

  $\begin{array}{l}S=10,000+5000\left(1\right)\\ S=15,000\\ S=10,000+5000\left(-1\right)\\ S=5,000\end{array}$

Hence, the maximum and minimum number of sheep are $15,000,5000$ .

c.

To determine

Graph both the equations.

Answer to Problem 41E

  

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the given equation, $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ for graph,

  

d.

To determine

Identify the wolves maximum population month.

Answer to Problem 41E

  $3and15$

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the graph of part $\left(c\right)$ , the wolves maximum population month is $3and15$ .

e.

To determine

Identify the sheep maximum population month.

Answer to Problem 41E

  $0,12and24$

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the graph of part $\left(c\right)$ , the sheep maximum population month is $0,12and24$ .

f.

To determine

Find the relationship between the wolves and sheep maximum population.

Answer to Problem 41E

When sheep population is at maximum then wolf population is increasing due to maximum availability of food it leads to maximum later.

Explanation of Solution

Given information:

The population of the wolves $W$ is modeled by $W=2000+1000\sin \left(\frac{\pi t}{6}\right)$ and population of the sheep $S$ is modeled by $S=10,000+5000\cos \left(\frac{\pi t}{6}\right)$ where $t$ is the time in months.

Calculation:

Consider the graph of part $\left(c\right)$ , the relationship between the wolves and sheep maximum population when sheep population is at maximum then wolf population is increasing due to maximum availability of food it leads to maximum later.