Chapter 4.2, Problem 56AYU

(a)

To determine

The number of insects were discovered. In other words, what was the population when $t=0?$

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

Answer to Problem 56AYU

The population when $t=0$ is $25$ .

Explanation of Solution

Given:

  $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

Calculation:

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

At the time they were discovered, $t=0,$ then the population is,

  $P\left(0\right)=\frac{50\left(1+0.5\left(0\right)\right)}{2+0.01\left(0\right)}=\frac{50}{2}=25$

Conclusion:

Hence,the population is $25$ .

(b)

To determine

Calculate the population be after $5$ years.

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

Answer to Problem 56AYU

The population after $5$ years is $596$ .

Explanation of Solution

Given:

  $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

Calculation:

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

After $5$ years, $t=60$ months, therefore the population is,

  $P\left(60\right)=\frac{50\left(1+0.5\left(60\right)\right)}{2+0.01\left(60\right)}=\frac{1550}{2.6}\approx 596$

Conclusion:

Hence, the population after $5$ years is $596$ .

(c)

To determine

The horizontal asymptote of $P\left(t\right)$ . What is the largest population that the protected area can sustain?

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

Answer to Problem 56AYU

The largest population that the protected area can sustain is $2500$ .

Explanation of Solution

Given:

  $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

The population $P$ of the insect $t$ months after being transplanted is $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}$

If $P\left(t\right)$ is a rational function with the degree of the numerator is same as the degree of the denominator, then $y=a/b$ is the horizontal asymptote.

Since $P\left(t\right)=\frac{50\left(1+0.5t\right)}{2+0.01t}=\frac{50+25t}{2+0.01t}$ has the degree of its numerator same as the degree of its denominator, therefore the horizontal asymptote is $y=\frac{25}{0.01}=2500$ which is the largest population of the insect can become.

Conclusion:

Hence, the largest population that the protected area can sustain is $2500$ .