Question

On a certain island there are two populations of deer. after t years the numbers of deer in the two populations are p(t)=100et and q(t)=1000e−t. when is the total population smallest?

Answer 1

t = .602 years = between 7 and 8 months

Answer 2

Total population of deer on the island will be least after 1.15 years.

  Two populations of deer have been given by two functions as,

$p(t)=100e^t$

$q(t)=1000e^{-t}$

If a function 'f' represents total population of deer after time 't',

Therefore, $f(t)=p(t)+q(t)$

To find the least population, find the derivative of function 'f' with respect to time 't' and equate it to zero to find the value of 't'.

$f(t)=100e^t+1000e^{-t}$

$f'(t)=100e^t-1000e^{-t}$

For $f'(t)=0$,

$100e^t-1000e^{-t}=0$

$100e^t=1000e^{-t}$

$100e^t=\frac{1000}{e^t}$

$100e^{2t}=1000$

$e^{2t}=10$

By taking natural log on both the sides of the equation,

$\text{ln}(e^{2t})=\text{ln}(10)$

$2t(\text{ln}e)=\text{ln}10$

$2t=\text{ln}10$

$t=\frac{\text{ln}10}{2}$

$t=1.151\approx 1.15$ years

   Therefore, total population of deer on the island will be least after 1.15 years.

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