1. Function Problem

Calculus function question?

The population of native cats in a large regional forest is modelled by the formula P=t^3-35t+275t+875 for 0 ≤ t ≤ 28, where t is the time in years after January 1980.

a) find the initial population of native cats in the forest

b) Use your calculator or other wise, to find the population when it was at its lowest level, and state the year in which this occurred.

All I need for question b) is the equation of the graph because the equation mentioned in the question does not work for some reason.... (to graph the function using my graphing calculator)

2. Originally Posted by dvmasaka
The population of native cats in a large regional forest is modelled by the formula P=t^3-35t+275t+875 for 0 ≤ t ≤ 28, where t is the time in years after January 1980.

a) find the initial population of native cats in the forest

b) Use your calculator or other wise, to find the population when it was at its lowest level, and state the year in which this occurred.

All I need for question b) is the equation of the graph because the equation mentioned in the question does not work for some reason.... (to graph the function using my graphing calculator)
I assume the 2nd term is -35t^2 ...

set xmin to 0 , xmax to 28 ... perform a zoom fit (zoom 0)

3. The global minimum will occur EITHER at one of the endpoints of the function, or where the derivative is $0$ at that point, and the second derivative is positive at that point.

$P = t^3 - 35t^2 + 275t +875$.

You should have found that the initial population is $875$ and the final population is $3087$.

$\frac{dP}{dt} = 3t^2 - 70t + 275$

$\frac{dP}{dt} = (t - 5)(3t - 55)$.

Setting the derivative equal to 0 to find the stationary points gives

$(t - 5)(3t - 55) = 0$

$t - 5 = 0$ or $3t - 55 = 0$

$t = 5$ or $t = \frac{55}{3}$.

Testing which of these is a minimum...

$\frac{d^2P}{dt^2} = 6t - 70$.

At $t = 5, \frac{d^2P}{dt^2} = -40$, so the function is at a maximum at $t = 5$.

At $t = \frac{55}{3}, \frac{d^2P}{dt^2} = 40$, so the function is at a minimum at $t = 40$.

When $t = \frac{55}{3}, P = \frac{8500}{27}$.

So the minimum population is $\approx 315$ cats, which occurred just after $18$ years.

4. thx

I finally get it now.

Thank you