91Ó°ÊÓ

The population P of the insect t months after being transplanted is

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

Substituting the value of t in the equation

$$\begin{gathered} P\left(t\right)=\frac{50(1+0.5t)}{2+0.01t} \\ t=0 \\ =\frac{50(1+0.5×0)}{2+0.01×0} \\ =25 \end{gathered}$$

The population P of the insect t months after being transplanted is

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

Substituting the value of t in the equation

$$\begin{gathered} P\left(t\right)=\frac{50(1+0.5t)}{2+0.01t} \\ t=5years \\ =5\left(12\right)months \\ =60 \\ P\left(t\right)=\frac{50(1+0.5t)}{2+0.01t} \\ =\frac{50(1+0.05×60)}{2+0.01×60} \\ =596.15 \end{gathered}$$

The population P of the insect t months after being transplanted is

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

Here, $n=1,m=1$

therefore, for finding horizontal asymptote we use the formula

role="math" localid="1646174992788" $$\begin{gathered} \frac{a_n}{b_m}i.e. \\ P=\frac{a_n}{b_m} \\ P\left(t\right)=\frac{50(1+0.5t)}{2+0.01t} \\ =\frac{25t+50}{0.01t+2} \\ therefore \\ {a}_n=25 \\ {b}_m=0.01 \\ =\frac{25}{0.01} \\ =2500 \end{gathered}$$

The largest population that the protected area can sustain is the population we get when t is large and this can be arbitrarily close to the horizontal asymptote

i.e. 2500 insect.