The average monthly rent for a 1000-sq-ft apartment in a major metropolitan area from 1998 through 2005 can be approximated by the function below where t is the time in years since the beginning of 1998. Find the value of t when rents were increasing most rapidly. Approximately when did this occur?

[tex]f(t) = 1.6859t^4 -18.553t^3+62.89t^2+6.5586t + 1009[/tex]

Recall that, we are asked for the value of t where it was increasing most rapidly. Then, this means that the first derivative has a positive value (this is because the function is increasing), but also, we want a value of t for which the first derivative is maximum. Then, let us calculate the derivative of f, and call it g. To do so, recall that the derivative of a polynomial [tex]ax^{n}=a\cdot nx^{n-1}[/tex]. Then

[tex]f'(t) = 6.7436t^3-55.659t^2+125.78 t + 6.5586= g(t)[/tex]

We want to find a value of t for which g is positive and maximum. For that, let us calculate the derivative of g. REcall that a function has a maximum/mininum where it's derivative is equal to zero. Then, we want to solve the following equation:

[tex]g'(t) = 20.2308 t^2 - 111.318 t + 125.78=0[/tex].

Recall the quadratic formula, that if we have a polynomial of the form [tex]ax^2+bx+c =0[/tex], the solution is given by [tex]x =\frac{-b\pm \sqrt[]{b^2-4ac}}{2}[/tex].

If we use this, we have the following solutions. t = 1.58851 and t = 3.91389. We want to evaluate which one gives us the maximum for g. Then, we are going to use the second derivative criteria. That is, a point is a maximum if and only if the first derivative at that point is zero and the second derivative is negative. The second derivative of g is

[tex]g''(t) = 40.4616 t - 111.318[/tex]

Note that g''(1.58851) =-47.0443<0 and g''(3.91389) = 47.0443>0, so t=1.58851 is the maximum we are looking for. Note that g(1.58851) = 92.94>0 as desired.

This means, that approximately one year and a halft later (that is at July-1999) the rents were increasing most rapidly.