MHB Mattie_j37890's Q: Estimate Critical Values of Function

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The discussion focuses on estimating the critical values of the function representing the average price of white sugar from 1993 to 2003. The first derivative of the function is set to zero to find the critical points, revealing four roots in the interval from 0 to 10 years. Using Newton's method, the approximate critical values are found at t ≈ 0.857, 4.607, 7.306, and 9.564. Evaluating the function at these critical points and the endpoints reveals that the absolute minimum price occurs at t ≈ 0.857 and the maximum at t ≈ 4.607. This analysis provides insight into the price fluctuations of sugar during the specified period.
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Here is the question:

What are the critical values of this function?

A model for the average price of a pound of white sugar in a certain country from August 1993 to August 2003 is given by the function
S(t) = −0.00003237t^5 + 0.0009037t^4 − 0.008956t^3 + 0.03629t^2 − 0.04467t + 0.4438
where t is measured in years since August of 1993. Estimate the times when sugar was cheapest and most expensive during the period 1993-2003. (Round your answers to three decimal places.)

I have posted a link there to this topic so the OP can see my work.
 
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Re: mattie_j37890's quetion at Yahoo! Answers regarding approximating critical values

Hello mattie_j37890,

We are given the function:

$$S(t)=-0.00003237t^5+0.0009037t^4-0.008956t^3+0.03629t^2-0.04467t+0.4438$$

First, let's look at a plot of this function just so we have some idea of what it looks like over the relevant domain:

View attachment 944

We can see that there are 4 relative extrema in the interval $0\le t\le10$

To find where they are, that is for what values of $t$ do they exist, we want to equate the first derivative to zero:

$$S'(t)=-0.00016185t^4+0.0036148t^3-0.026868t^2+0.07258t-0.04467=0$$

Let's look at a plot of the derivative to get some idea of where the roots are:

View attachment 945

We may use Newton's method to approximate these roots. We may write:

$$f(t)=-0.00016185t^4+0.0036148t^3-0.026868t^2+0.07258t-0.04467=0$$

$$f'(t)=-0.0006474t^3+0.0108444t^2-0.053736t+0.07258$$

And so we have the recursion:

$$t_{n+1}=t_{n}-\frac{f\left(t_{n} \right)}{f'\left(t_{n} \right)}$$

We see the first root is near $t=1$, and so we may compute:

$$t_0=1$$

$$t_1\approx0.845220550257$$

$$t_2\approx0.857337077417$$

$$t_3\approx0.857416179268$$

$$t_4\approx0.857416182626$$

$$t_5\approx0.857416182626$$

We see the second root is near $t=5$, and so we may compute:

$$t_0=5$$

$$t_1\approx4.53064243449$$

$$t_2\approx4.60497652706$$

$$t_3\approx4.60659458701$$

$$t_4\approx4.606595386$$

$$t_5\approx4.606595386$$

We see the third root is near $t=7$, and so we may compute:

$$t_0=7$$

$$t_1\approx7.3250339402$$

$$t_2\approx7.30609727792$$

$$t_3\approx7.30607003007$$

$$t_4\approx7.30607003001$$

$$t_5\approx7.30607003001$$

We see the fourth root is near $t=10$, and so we may compute:

$$t_0=10$$

$$t_1\approx9.66222062004$$

$$t_2\approx9.57077528295$$

$$t_3\approx9.56421133076$$

$$t_4\approx9.56417851958$$

$$t_5\approx9.56417851876$$

$$t_6\approx9.56417851876$$

And so, we have the four roots of the first derivative approximated by:

$$t\approx0.857416182626,\,4.606595386,\,7.30607003001,\,9.56417851876$$

Now, to find the absolute extrema on the given closed interval, we need to evaluate the function at the endpoints and at the critical values:

$$S(0)=0.4438$$

$$S(0.857416182626)=0.4270063563925$$

$$S(4.606595386)=0.47243084424924303$$

$$S(7.30607003001)=0.462862495347419$$

$$S(9.56417851876)=0.47195253062506465$$

$$S(10)=0.4701$$

And so rounding to 3 decimal places, we find:

The absolute minimum occurs when $$t\approx0.857$$

The absolute maximum occurs when $$t\approx4.607$$
 

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