# A rare species of insect was discovered in the amazon rainforest. To protect the species, environmentalists declare theinsect endangered and transplant the insect into a protected area.The population P of the insect t months after being transplantedis P(t)=(50(1+0.5t))/((2+0.01t)) (a) How many insects were discovered? In other words, what wasthe population when t=0? (b) What will the population be after 5 years? (c)Determine the horizontal asymptote of P(t).What is thelargest population that the protected area can sustain?

A rare species of insect was discovered in the amazon rainforest. To protect the species, environmentalists declare theinsect endangered and transplant the insect into a protected area.The population P of the insect t months after being transplantedis
$P\left(t\right)=\frac{50\left(1+0.5t\right)}{\left(2+0.01t\right)}$
(a) How many insects were discovered? In other words, what wasthe population when t=0?
(b) What will the population be after 5 years?
(c)Determine the horizontal asymptote of P(t).What is thelargest population that the protected area can sustain?
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grocbyntza
a.)
P(0) is the population at t=0 so substitute into the equation and solve
$P\left(0\right)=\frac{50\left(1+.5\left(0\right)\right)}{2+.01\left(0\right)}=25$
b.)
because t is in months convert years to months and substitute (60months)
$P\left(0\right)=\frac{50\left(1+.5\left(60\right)\right)}{2+.01\left(60\right)}=596.1$
c.)
You basically want to know the value when T equals $\mathrm{\infty }$. This will result in the ratio of the coefficents of the variable which will be 25/.01 which is 2500. So 2500 is the max this population can handle.