Answer
The carrying capacity is $298.73$ and the number of animals in the sanctuary after 20 years is $221$
Work Step by Step
Follow the Logistic Growth Models:
$P(t)=\frac{CP_0}{P_0+(C-P_0)e^{-rt}}$
$r=\frac{1}{t_1} \ln[\frac{P_2(P_1-P_0)}{P_0(P_2-P_1)}]$
$C=\frac{P_0[P_1(P_0+P_2)-2P_0P_2]}{P^2_1-P_0P_2}$
From the given problem we have:
$t_1=5$
$P_0=500$
$P_1=800$
$P_2=800$
Substitute and solve for $r$:
$r=\frac{1}{2} \ln[\frac{76(62-50)}{50(76-62)}]$
$r=0.132$
Solve for C:
$C=\frac{62[62(50+76)-2.50.76]}{62^2-50.76}$
$C=298.73$
Thus,
$P(t)=\frac{298.73\times500}{50+(298.73-50)e^{-0.132\times20}}$
$P(t)=221$
The carrying capacity is $298.73$ and the number of animals in the sanctuary after 20 years is $221$
