Find the absolute maximum/minimum values of S(t) where S'(t) is a quartic function with lots of horrible decimal places.

Question

So I have a problem where I'm to find the absolute maximum and minimum values of the following function...

$S(t) = -0.00003237t^5 + 0.0009037t^4 - 0.008956t^3 + 0.03629t^2 -0.04458t + 0.4074$

...within the interval (1993,2003)

Differentiating this and setting S'(t) = 0 gives...

$S'(t) = -5(0.00003237)t^4 + 4(0.0009037)t^3 - 3(0.008956)t^2 + 2(0.03629)t - 0.04458 = 0$

Now I'm at a dead end. I have no idea how to factorise this and find the values of t where S'(t) = 0.

Any ideas?

Answer 1

This is a polynomial of degree 4. Theoretically you could solve this analitically with a super complicated procedure (see https://en.wikipedia.org/wiki/Quartic_function). By calculating the value for $S'(t)$ at the positions $t = 1,3,6,9,10$ you will see that the signs change each time and therfore the zeros must be in between. As the coefficient of $t^4$ is negative, the curve will go to $-\inf$ for $x \to \pm \inf$ So the absolute minimum and maximum within the interval $(1993,2003)$ do not exist (because it does not contain 1993 and 2003) but the suppremum / infimum are at the given positions, all because the function is monotonically decreasing within this interval.

For iteratively approximating the zeros of $S'(t)$ you could use for example Newton's Method (https://en.wikipedia.org/wiki/Newton%27s_method)

Answer 2

If we take that derivative and find the zeros (using some numerical method), we find four points at:

$$ t = 0.854778, 4.61772, 7.29191, 9.56986 $$

These are well outside the range of interest.

Next, trying the endpoints (plot it), we find:

• Global = Local max at $t = 1993$
• Global min at $t = 2003$

One thing I always do before the math is do a plot and see what strikes you about it and then the analytical part.

Answer 3

Another way to see that the function is strictly decreasing on the interval is to look at the rate of change of just the $t^5$ term. It changes at a rate of $(-5) 0.00003237t^4$. For $t$-values near $2000$, $(-5)0.00003237t^4\approx(-5)0.00003237\cdot2000t^3=-0.3237t^3$. The rate of change of the remaining terms of the polynomial is much, much smaller in absolute value, so they can't pull the large negative rate of change due to the $t^5$ term anywhere near zero.

Once you see that the function is decreasing on the interval, you know immediately that the absolute maximum is at $t=1993$ and the absolute minimum is at $t=2003$.

Answer 4

A somewhat homelier approach from the "theory of equations" is to start by dividing $ \ S \ '(t) \ $ through by its leading coefficient to produce a monic polynomial. We have $ \ [ \ -5(0.00003237) \ ]^{-1} \ \approx \ -6180 \ $ , so our polynomial becomes roughly

$$ S \ '(t) \ \ \rightarrow \ \ t^4 \ - \ 22t^3 \ + \ 166 t^2 \ - \ 449 t \ + \ 276 \ = \ 0 \ \ . $$

[This will be sufficient precision for our demonstration.]

We have a theorem which states that if $ \ c \ > \ 0 \ $ and division of a polynomial $ \ f(x) \ $ by $ \ (x-c) \ $ produces a quotient polynomial with only positive coefficients and a positive remainder, then $ \ c \ $ is an upper bound on the real zeroes of the polynomial. Synthetic division by, say, $ \ t \ = \ 30 \ $ , on $ \ S \ '(t) \ $ yields

$$ 30 \ \ / \ \begin{array}{c} \ 1&-22& \ 166&\ \ -449&\ \ \ \ \ 276 \end{array} $$ $$ \quad \quad \quad \quad \quad \quad \ \begin{array}{c} \ 30& \ 240&\ \ \ 12180&351930 \end{array} \\ $$ $$ \quad \quad \quad \ \ \ \begin{array}{c} 1& \ \ \ \ \ 8& \ \ \ 406& \ \ \ \ 11731&352206 \end{array} $$

We may conclude from this that $ \ S \ '(t) \ $ has no zeroes larger than 30. There are then no local extrema in the interval in question. We see fairly easily that $ \ S \ '(t) \ < \ 0 $ in that interval, so the function $ \ S(t) \ $ is strictly decreasing and the absolute extrema on the closed interval would lie at its endpoints. (As flawr points out, there are no extrema in the open interval.)